UC-NRLF 


EH  MEMOEIAM 
Edward  Bright 


(Brcenlcafs  Jttathematical  3ttk^ 


UNIVERSITY  ALGEBRA. 


DESIGNED   FOR  THE   USE   OF   SCHOOLS 
AND  COLLEGES. 


PREPARED   BY 


WEBSTER  WELLS,  S.  B., 

i 

ASSISTANT   PROFESSOR   OF   MATHEMATICS    IN   THE   MASSACHUSETTS 
INSTITUTE   OF   TECHNOLOGY; 


LEACH,    SHEWELL,    AND    SANBORN, 
BOSTON    AND    NEW    YORK. 


COPYRIGHT,  1880. 


q/v- 


By  WEBSTER  WELL?.. 


PREFACE. 


This  work  was  designed  to  take  the  place  of  Green- 
leaf's  Higher  Algebra,  portions  of  which  have  been  used 
in  the  preparation  of  the  present  volume.  It  contains 
the  topics  usually  taught  in  High  Schools  and  Colleges, 
and  the  author's  aim  has  been  to  present  the  subject  in  a 
compact  form  and  in  clear  and  concise  language.  The 
principles  have  been  developed  with  regard  to  logical  ac- 
curacy, and  care  has  been  given  to  the  selection  of  exam- 
ples and  practical  illustrations  which  should  exercise  the 
student  in  all  the  common  applications  of  the  algebraic 
analysis.  The  full  treatment  given  in  the  earlier  chap- 
ters renders  the  previous  study  of  a  more  elementary 
text-book  unnecessary. 

Attention  is  invited  to  the  following  chapters,  including 
those  in  which  the  most  important  changes  have  been 
made  in  the  Higher  Algebra  :  — 

Parentheses. 

Factoring. 

Zero  and  Infinity. 

Theory  of  Exponents. 

Simultaneous  Equations  involving  Quadratics. 

Binomial  Theorem  for  Positive  Integral  Exponents. 

Undetermined  Coefficients. 

Logarithms. 

The  answers  have  been  put  by  themselves  in  the  back 
part  of  the  book,  and  those  have  been  omitted  which,  if 


7979-5 


i  /  PREFACE. 

given,  would  destroy  the  utility  of  the  problem.  The  ex- 
amples are  over  eighteen  hundred  in  number,  and  are  pro- 
gressive, commencing  with  simple  applications  of  the 
rules,  and  passing  gradually  to  those  which  require  some 
thought  for  their  solution. 

The  works  of  Todhunter  and  Hamblin  Smith,  and  other 
standard  volumes,  have  been  consulted  in  the  preparation 
of  the  work,  and  have  furnished  a  number  of  examples 
and  problems.  The  author  has  also  received  numerous 
suggestions  from  practical  teachers,  to  whom  he  would 
here  express  his  thanks. 

WEBSTER  WELLS. 

Boston,  1884. 


UNIVERSITY  ALGEBRA. 


CONTENTS. 


Chapter  Paob 

I.    Definitions  and  Notation 1 

Symbols  of  Quantity 1 

Symbols  of  Operation 2 

Symbols  of  Relation 4 

Symbols  of  Abbreviation 5 

Algebraic  Expressions 5 

Axioms 8 

Negative  Quantities 12 

II.    Addition 14 

III.  Subtraction 19 

IV.  Use  of  Parentheses 21 

V.    Multiplication 24 

VI.    Division 31 

VII.    Formula 38 

VIII.    Factoring 40 

IX.    Greatest  Common  Divisor 53 

X.    Least  Common  Multiple 61 

XI.    Fractions 66 

Reduction  of  Fractions 70 

Addition  and  Subtraction  of  Fractions 78 

Multiplication  of  Fractions 82 

Division  of  Fractions  85 

Complex  Fractions  87 

XII.    Simple  Equations.  —  One  unknown  quantity 89 

Transformation  of  Equations 91 

Solution  of  Equations  95 


Vlll 


CONTENTS. 


XIII.  Problems. — One  unknown  quantity 103 

XIV.  Simple  Equations.  —  Two  unknown  quantities 113 

Elimination 115 

XV.    Simple  Equations.  —  More  than  two  unknown  quan- 
tities   123 

XVI.    Problems. —More  than  one  unknown  quantity 127 

Generalization  of  Problems 133 

XVII.    Discussion  of  Problems 137 

Interpretation  of  Negative  Results  139 

XVIII.    Zero  and  Infinity 142 

Problem  of  the  Couriers 143 

XIX.    Inequalities  148 

XX.    Involution 153 

Involution  of  Monomials 153 

Involution  of  Polynomials 154 

Square  of  a  Polynomial  155 

Cube  of  a  Binomial 156 

Cube  of  a  Polynomial 157 

XXI.    Evolution 158 

Evolution  of  Monomials 159 

Square  Root  of  Polynomials 160 

Square  Root  of  Numbers 163 

Cube  Root  of  Polynomials 168 

Cube  Root  of  Numbers 171 

Any  Root  of  Polynomials 174 

XXII.    The  Theory  of  Exponents 176 

XXIII.    Radicals 188 

Reduction  of  Radicals 1 88 

Addition  and  Subtraction  of  Radicals  193 

Multiplication  of  Radicals 194 

Division  of  Radicals 196 

Involution  of  Radicals 197 

Evolution  of  Radicals 198 

Reduction  of  Fractions  with  Irrational  Denominators  199 

Imaginary  Quantities 202 

Quadratic  Surds 205 

Radical  Equations 208 


CONTENTS.  ix 

XXIV.     Quadratic  Equations.  —  One  unknown  quantity...  210 

Pure  Quadratic  Equations 211 

Affected  Quadratic  Equations  213 

XXV.    Problems.  —  Quadratic  Equations.  —  One  unknown 

quantity 223 

XXVI.     Equations  in  the  Quadratic  Form 227 

XXVII.    Simultaneous  Equations  involving  Quadratics 233 

XXVIII.    Problems. —  Quadratic  Equations.  —  Two  unknown 

QUANTITIES 244 

XXIX.    Theory  of  Quadratic  Equations 249 

Discussion  of  the  General  Equation  249 

XXX.    Discussion    of    Problems    leading    to    Quadratic 

Equations 257 

Interpretation  of  I maginary  Results 259 

Problem  of  the  Lights  259 

XXXI.  Ratio  and  Proportion 262 

XXXII.  Variation 270 

XXXIII.  Arithmetical  Progression 274 

XXXIV.  Geometrical  Progression 282 

XXXV.  Harmonical  Progression 291 

XXXVI.    Permutations  and  Combinations 294 

XXXVII.    Binomial  Theorem.  —  Positive  Integral  Exponent  298 

XXXVIII.    Undetermined  Coefficients  304 

Expansion  of  Fractions  into  Series 307 

Expansion  of  Radicals  into  Series 310 

Decomposition  of  Rational  Fractions 312 

Reversion  of  Series 318 

XXXIX.    Binomial  Theorem. —Any  Exponent 321 

XL.    Summation  of  Infinite  Series 328 

Recurring  Series 328 

The  Differential  Method 332 

Interpolation  336 


x  CONTENTS. 

XLI.     Logarithms 339 

Properties  of  Logarithms 342 

Use  of  the  Table 348 

Solutions  of  Arithmetical  Problems  by  Logarithms 354 

Exponential  Equations 358 

Application  of  Logarithms  to  Problems  in  Compound  In- 
terest    359 

Exponential  and  Logarithmic  Series 362 

Arithmetical  Complement 366 

XLII.     General  Theory  of  Equations 369 

Divisibility  of  Equations 370 

Number  of  Roots .- 371 

Formation  of  Equations 373 

Composition  of  Coefficients 374 

Fractional  Roots 376 

Imaginary  Roots 376 

Transformation  of  Equations 377 

Descartes'  Rule  of  Signs 383 

Derived  Polynomials 385 

Equal  Roots 386 

Limits  of  the  Roots  of  an  Equation 389 

Sturm's  Theorem 392 

XLIII.     Solution  of  Higher  Numerical  Equations 399 

Commensurable  Roots 400 

Recurring  or  Reciprocal  Equations 404 

Cardan's  Method  for  the  Solution  of  Cubic  Equations. ....  408 

Biquadratic  Equations 411 

Incommensurable  Roots 412 

Horner's  Method 412 

Approximation  by  Double  Position 417 

Newton's  Method  of  Approximation 419 

Answers  to  Examples 421 

Table  of  the  Logarithms  of  Numbers  from  1  to  10,000.  .Appendix. 


ALGEBRA. 


I.  —  DEFINITIONS   AND    NOTATION. 

1.  Quantity  is  anything  that  can  be  measured ;  as  dis- 
tance, time,  weight,  and  number. 

2.  The  Measurement  of  quantity  is  accomplished  by  find- 
ing the  number  of  times  it  contains  another  quantity  of  the 
same  kind,  assumed  as  a  standard.  This  standard  is  called 
the  unit  of  measure. 

3.  Mathematics  is  the  science  of  quantities  and  their  re- 
lations. 

4.  Algebra  is  that  branch  of  mathematics  in  which  the 
relations  of  quantities  are  investigated,  and  the  reasoning 
abridged  and  generalized,  by  means  of  symbols. 

5.  The  Symbols  employed  in  Algebra  are  of  four  kinds: 
symbols  of  quantity,  symbols  of  operation,  symbols  of  relation, 
and  symbols  of  abbreviation. 

SYMBOLS   OF  QUANTITY. 

6.  The  Symbols  of  Quantity  generally  used  are  the 
figures  of  Arithmetic  and  the  letters  of  the  alphabet. 

The  figures  are  used  to  represent  known  quantities  and 
determined  values,  and  the  letters  any  quantities  whatever, 
known  or  unknown. 

7.  Known   Quantities,   or  those  whose  values  are  given, 


■2  ALGEBRA. 

when  not  expressed  by  figures,  are  usually  represented  by  the 
first  letters  of  the  alphabet,  as  a,  b,  c. 

8.  Unknown  Quantities,  or  those  whose  values  are  not 
given,  are  usually  represented  by  the  last  letters  of  the 
alphabet,  as  x,  y,  z. 

9.  Zero,  or  the  absence  of  quantity,  is  represented  by  the 
symbol  0. 

10.  Quantities  occupying  similar  relations  in  different  op- 
erations are  often  represented  by  the  same  letter,  distinguished 
by  different  accents,  as  a',  a",  a'",  read  "a  prime,"  "a  second," 
"a  third,"  etc. ;  or  by  different  subscript  figures,  as  a1}  a2)  as, 
read  "  a  one/'  "  a  two,"  "  a  three,"  etc. 

SYMBOLS   OF  OPERATION. 

11.  The  Symbols  of  Operation  are  certain  signs  or  char- 
acters used  to  indicate  algebraic  operations. 

12.  The  Sign  of  Addition,  +  ,  is  called  "plus."  Thus, 
a  -(-  b,  read  "  a  plus  b"  indicates  that  the  quantity  b  is  to  be 
added  to  the  quantity  a. 

13.  The  Sign  of  Subtraction,  — ,  is  called  "minus." 
Thus,  a  —  b,  read  "  a  minus  b"  indicates  that  the  quantity 
b  is  to  be  subtracted  from  the  quantity  a. 

The  sign  ~  indicates  the  difference  of  two  quantities  when 
it  is  not  known  which  of  them  is  the  greater.  Thus,  a  ~  b 
indicates  the  difference  of  the  two  quantities  a  and  b. 

14.  The  Sign  of  Multiplication,  x ,  is  read  "  times" 
"into,"  or  "multiplied  by."  Thus,  aXb  indicates  that  the 
quantity  a  is  multiplied  by  the  quantity  b. 

A  simple  point  (. )  is  sometimes  used  in  place  of  the  sign  X- 
The  sign  of  multiplication  is,  however,  usually  omitted,  except 
between  two  arithmetical  figures  separated  by  no  other  sign; 
multiplication  is  therefore  indicated  by  the  absence  of  any 
sign.     Thus,  2ab  indicates  the  same  as  2  X  a  X  b,  or  2  .  a  .  b. 


DEFINITIONS   AND    NOTATION.  3 

15.  The  quantities  multiplied  are  called  factors,  and  the 
result  of  the  multiplication  is  called  the  product. 

16.  The  Sign  of  Division,  ~,  is  read  "divided  by." 
Thus,  a  -4-  b  indicates  that  the  quantity  a  is  divided  by  the 
quantity  b. 

Division  is  otherwise  often  indicated  by  writing  the  divi- 
dend above,  and  the  divisor  below,  a  horizontal  line.     Thus, 

-   indicates  the  same  as  a  -i-  b.      Also,  the  sign  of  division 

b 

may  be  replaced  in  an  operation  by  a  straight  or  curved  line. 

Thus,  a  I  b,  or  b )  a,  indicates  the  same  as  a  -f-  b. 

17.  The  Exponential  Sign  is  a  figure  or  letter  written  at 
the  right  of  and  above  a  quantity,  to  indicate  the  number  of 
times  the  quantity  is  taken  as  a  factor.  Thus,  in  x3,  the  3  in- 
dicates that  x  is  taken  three  times  as  a  factor ;  that  is,  x3  is 
equivalent  to  xxx. 

The  product  obtained  by  taking  a  factor  two  or  more  times 
is  called  a  power.  A  single  letter  is  also  often  called  the 
first  power  of  that  letter.     Thus, 

a2  is  read  "a  to  the  second  power,"  or  "a  square,"  and 
indicates  a  a; 

a3  is  read  "  a  to  the  third  power,"  or  "  a  cube,"  and  indi- 
cates acta; 

a4  is  read  "  a  to  the  fourth  power,"  or  "  a  fourth,"  and  indi- 
cates a  a  a  a; 

an  is  read  "  a  to  the  nth  power,"  or  "  a  nth,"  and  indicates 
a  a  a  etc.,  to  n  factors. 

The  figures  or  letters  used  to  indicate  powers  are  called 
exponents ;  and  when  no  exponent  is  written,  the  first  power 
is  understood.     Thus,  a  is  equivalent  to  a1. 

The  root  of  a  quantity  is  one  of  its  equal  factors.  Thus, 
the  root  of  a2,  a3,  or  a*  is  a. 

18.  The  Radical  Sign,  \f ,  when  prefixed  to  a  quantity, 
indicates  that  some  root  of  the  quantity  is  to  be  extracted. 


4  ALGEBRA. 

Thus, 

SJ a  indicates  the  second  or  square  root  of  a; 

^]a  indicates  the  third  or  cuhe  root  of  a; 

$  a  indicates  the  fourth  root  of  a;    and  so  on. 

The  index  of  the  root  is  the  figure  or  letter  written  over 
the  radical  sign.  Thus,  2  is  the  index  of  the  square  root,  3  of 
the  cube  root ;  and  so  on. 

When  the  radical  sign  has  no  index  written  over  it,  the 
index  2  is  understood.     Thus,  sj  a  is  the  same  as  tf  a. 

SYMBOLS   OF  RELATION. 

19.  The  Symbols  of  Relation  are  signs  used  to  indicate 
the  relative  magnitudes  of  quantities. 

20.  The  Sign  of  Equality,  =,  read  " equals"  or  "equal 
to,"  indicates  that  the  quantities  between  which  it  is  placed 
are  equal.  Thus,  x  =  y  indicates  that  the  quantity  x  is  equal 
to  the  quantity  y. 

A  statement  that  two  quantities  are  equal  is  called  an 
equation.  Thus,  x  -\-  4=2  x —  1  is  an  equation,  and  is  read 
"  x  plus  4  equals  2x  minus  1." 

21.  The  Sign  of  Ratio,  : ,  read  "  to,"  indicates  that  the 
two  quantities  between  which  it  is  placed  are  taken  as  the 
terms  of  a  ratio.  Thus,  a  :  b  indicates  the  ratio  of  the  quan- 
tity a  to  the  quantity  b,  and  is  read  "  the  ratio  of  a  to  b." 

A  proportion,  or  an  equality  of  ratios,  is  expressed  by  writ- 
ing the  sign  =,  or  the  sign  :  :,  between  equal  ratios.  Thus, 
30  :  6  =  25  :  5  indicates  that  the  ratio  of  30  to  G  is  equal  to 
the  ratio  of  25  to  5,  and  is  read  "30  is  to  6  as  25  is  to  5." 

22.  The  Sign  of  Inequality,  >  or  <  ,  read  "  is  greater 
than"  or  "is  less  than"  respectively,  when  placed  between 

two  quantities,  indicates  that  the  quantity  toward  which  the 
opening  of  the  sign  turns  is  the  greater.  Tims,  x  >  y  is 
read  "x  is  greater  than  y" ;  x—6<  y  is  read  "x  minus  6 
is  less  than  y." 


DEFINITIONS   AND   NOTATION.  5 

23.     The  Sign  of  Variation,  cc,  read  "varies  as,"  indicates 
that   the  two  quantities  between  which  it  is  placed  increase 

or  diminish  together,  in  the  same  ratio.     Thus,  a  oc  _  is  read 
"  a  varies  as  c  divided  by  d." 


SYMBOLS   OF  ABBREVIATION. 

24.  The  Signs  of  Deduction,  .-.  and  v ,  stand  the  one  for 
therefore  or  hence,  the  other  for  since  or  because. 

25.  The  Signs  of  Aggregation,  the  vinculum  ,  the 

bar  | ,  the  parenthesis  (  )  ,  the  brackets  [  ]  ,  and  the  braces  j  £, 

indicate  that  the  quantities  connected  or  enclosed  by  them  are 
to  be  subjected  to  the  same  operations.     Thus, 

a  +  b  X  oc,     a  x,     (a  +  b)  x,    [a  +  fr]x,    \a  +  bc  x, 


b 


all    indicate    that    the    quantity    a  +  b   is    to    be    multiplied 
by  x. 

26.  The  Sign  of  Continuation,    ,  stands  for  and  so 

on,  or  continued  by  the  some  law.     Thus, 

a,  a  +  b,    a  +  2  b,    a  +  3  b,  is  read 

"a,  a  plus  b,  a  plus  2b,  a  plus  3  b,  and  so  on." 

ALGEBRAIC  EXPRESSIONS. 

27.  An  Algebraic  Expression  is  any  combination  of  alge- 
braic symbols. 

28.  A  Coefficient  of  a  quantity  is  a  figure  or  letter  pre- 
fixed to  it,  to  show  how  many  times  the  quantity  is  to  be 
taken.  Thus,  in  4«,  4  is  the  coefficient  of  a,  and  indicates 
that  a  is  taken  four  times,  or  a  +  a  +  a  +  a.  Where  any 
number  of  quantities  are  multiplied  together,  the  product  of 


6  ALGEBEA. 

any  of  them  may  be  regarded  as  the  coefficient  of  the  product 
of  the  others;  thus,  in  abed,  ab  is  the  coefficient  of  ed, 
b  of  ac  d,  a  b  d  of  c,  and  so  on. 

When  no  coefficient  of  a  quantity  is  written,  1  is  understood 
to  he  the  coefficient.  Thus,  a  is  the  same  as  1  a,  and  x  y  is 
the  same  as  1  x  y. 

29.  The  Terms  of  an  algebraic  expression  are  its  parts 
connected  by  the  signs  +  or  — .     Thus, 

a  and  b  are  the  terms  of  the  expression  a  +  b ; 

2  a,   b2,   and  —  2  a  c,   of  the  expression  2  a  +  b2  —  2  a  c. 

30.  The  Degree  .of  a  term  is  the  number  of  literal  factors 
which  it  contains.     Thus, 

2  a  is  of  the  first  degree,  as  it  contains  but  one  literal  factor, 
a  &  is  of  the  second  degree,  as  it  contains  two  literal  factors. 

3  a  b2  is  of  the  third  degree,  as  it  contains  three  literal  factors. 

The  degree  of  any  term  is  determined  by  adding  the  expon- 
ents of  its  several  letters.     Thus,  a  b2  cs  is  of  the  sixth  degree. 

31.  Positive  Terms  are  those  preceded  by  a  plus  sign  ;  as, 

+  2  a,  or  +  a  b2. 

When  a  term  has  no  sign  written,  it  is  understood  to  be 
positive.     Thus,  a  is  the  same  as  +  a. 

Negative  Terms  are  those  preceded  by  a  minus  sign  ;  as, 

—  3  a,  or  —be. 

This  sign  can  never  be  omitted. 

32.  In  a  positive  term,  the  coefficient  indicates  how  many 
times  the  quantity  is  taken  additively  (Art.  28)  ;  in  a  nega- 
tive term,  the  coefficient  indicates  how  many  times  the  quan- 
tity is  taken  svbtractively.     Thus, 

+  2  x  is  the  same  as  +  x  +  x ; 

—  3  a  is  the  same  as  —  a  —  a  —  a. 


DEFINITIONS  AND   NOTATION.  7 

33.  If  the  same  quantity  be  both  added  to  and  subtracted 

from   another,   the  value   of  the  latter  will  not  be  changed  ; 

hence  if  any  quantity  b  be  added  to  any  other  quantity  a,  and 

b  be   subtracted  from   the   result,   the  remainder  will  be  a; 

that  is, 

#       a  +  b  —  b  =  a. 

Consequently,  equal  terms  affected  by  unlike  signs,  in  an 
expression,  neutralize  each  other,  or  cancel. 

34.  Similar  or  Like  Terms  are  those  which  differ  only  in 
their  numerical  coefficients.     Thus, 

2  x  y2  and  —  1  xif  are  similar  terms. 

Dissimilar  or  Unlike  Terms  are  those  which  are  not  similar. 
Thus, 

b  x2y  and  bxy2  are  dissimilar  terms. 

35.  A  Monomial  is  an  algebraic  expression  consisting  of 
only  one  term  ;  as,  5  a,  7  a  b,  or  3  b'2  c. 

A  monomial  is  sometimes  called  a  simple  quantity. 

36.  A  Polynomial  is  an  algebraic  expression  consisting  of 
more  than  one  term  ;  as,  a  +  b,  or  3  a2  +  b  —  5  b3. 

A  polynomial  is  sometimes  called  a  compound  quantity,  or  a 
multinomial. 

37.  A  Binomial  is  a  polynomial  of  two  terms ;  as, 

a  —  b,  2 a  +  b2,  ov  dad2  —  b. 

A  binomial  whose  second  term  is  negative,  as  a  —  b,  is  some- 
times called  a  residual. 

38.  A  Trinomial  is  a  polynomial  of  three  terms ;  as, 

a  +  b  +  c,  or  a  b  +  c2  —  b3. 

39.  Homogeneous  Terms  are  those  of  the  same  degree ;  as, 

a2,  3  be,  and  —  4 x2. 


$  ALGEBRA. 

40.  A  polynomial  is  homogeneous  when  all  its  terms  are 
homogeneous  ;  as,  a3  +  2  a  b  c  —  3  b3. 

41.  A  polynomial  is  said  to  be  arranged  according  to  the 
decreasing  powers  of  any  letter,  when  the  term  having  the 
highest  exponent  of  that  letter  is  placed  first,  that  having 
the  next  lower  immediately  after,  and  so  on.     Thus, 

a3+3a2b  +  3ab2  +  b3 

is  arranged  according  t6  the  decreasing  powers  of  a. 

A  polynomial  is  said  to  be  arranged  according  to  the  increas- 
ing powers  of  any  letter,  when  the  term  having  the  lowest 
exponent  of  that  letter  is  placed  first,  that  having  the  next 
higher  immediately  after,  and  so  on.     Thus, 

a3  +  3  a2  b  +  3  a  b2  +  b3 

is  arranged  according  to  the  increasing  powers  of  b. 

42.  The  Reciprocal  of  a  quantity  is  1  divided  by  that 
quantity.     Thus,  the  reciprocal  of 

a  is  - ,  and  of  x  +  y  is 


a  x  +  y 

43.  The  Interpretation  of  an  algebraic  expression  consists 
in  rendering  it  into  an  arithmetical  quantity,  by  means  of  the 
numerical  values  assigned  to  its  letters.  The  result  is  called 
the  numerical  value  of  the  expression. 

Thus,  the  numerical  value  of 

4d+  3  be  —  d 
when  a  =  4,  b  =  3,  c  =  5,  and  d  =  2,  is 

4x4  +  3x3x5-2  =  16  +  45  -  2  =  59. 

AXIOMS. 

44.  An  Axiom  is  a  self-evident  truth. 

Algebraic  operations  are  based  upon  definitions,  and  the 
following  axioms :  — 


DEFINITIONS  AND   NOTATION.  9 

1.  If  the  same  quantity,  or  equal  quantities,  be   added  to 
equal  quantities,  the  sums  will  he  equal. 

•   2.  If  the  same  quantity,  or  equal  quantities,  he  subtracted 
from  equal  quantities,  the  remainders  will  he  equal. 

3.  If  equal  quantities  be  multiplied  by  the  same  quantity,  or 
by  equal  quantities,  the  products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  the  same  quantity,  or 
by  equal  quantities,  the  quotients  will  be  equal. 

5.  If  the  same  quantity  be  both  added  to  and  subtracted  from 
another,  the  value  of  the  latter  will  not  be  changed. 

6.  If  a  quantity  be  both  multiplied  and  divided  by  another, 
the  value  of  the  former  will  not  be  changed. 

7.  Quantities  which  are  equal  to  the  same  quantity  are  equal 
to  each  other. 

8.  Like  powers  and  like  roots  of  equal  quantities  are  equal. 

9.  The  whole  of  a  quantity  is  equal  to  the  sum  of  all  its 
parts. 

EXERCISES  ON  THE   PRECEDING  DEFINITIONS  AND 

PRINCIPLES. 

45.     Translate    the   following   algebraic   expressions    into 
ordinary  language : 


d 


m 


1.  3  a2  +  b  c  —  q.  5.  cd  :  —  =  ab  :  \J  x*. 


3 


n 


x 


2.  4  m .  6.  (a  —  b)x  =  [c  +  d~]  y. 

<-  -}         3a  —  d 

"  2c  +  b 


3.  ^  a  +  b  =  ^  a2  —  c.  7.    {m  +  r  —  s}n  = 


4,mn>pa.  8.  \J  -^-  <  (e  -  d)  (h  +  fj. 

46.     Put  into  the  form  of  algebraic  expressions  the  follow- 
ing: 

1.  Five  times  a,  added  to  two  times  b. 

2.  Two  times  x,  minus  y  to  the  second  power. 


10  ALGEBRA. 

3.  The  difference  of  x  and  y. 

4.  The  product  of  «,  b,  c  square,  and  d  cube. 

5.  x  +  y  multiplied  by  a  —  b. 

6.  a  square  divided  by  the  sum  of  b  and  c. 

7.  x  divided  by  3,  increased  by  2,  equals  three  times  y, 
diminished  by  11. 

8.  The  reciprocal  of  a  +  b,  plus  the  square  of  a}  minus  the 
cube  root  of  b,  is  equal  to  the  square  root  of  c. 

9.  The  ratio  of  5  a  divided  by  b,  to  d  divided  by  c  square, 
equals  the  ratio  of  x  square  y  cube  to  y  square  z  fourth. 

10.  The  product  of  in  and  a  +  b  is  less  than  the  reciprocal 
of  x  cube. 

11.  The  product  of  x  +  y  and  x  —  y  is   greater  than  the 
product  of  the  square  of  a  —  d  into  the  cube  of  a  +  b. 

12.  The  quotient  of  a  divided  by  3  a  — 2  is  equal  to  the 
square  root  of  the  quotient  of  m  +  n  divided  by  2x  —  y2. 

47.     Find  the  numerical  values  of  the  following :  — 
When  a  =  6,  b  =  5,  c  =  4,  and  d  —  1,  of 

1.  a2  +  2 a b  —  c  +  d.  4.  a2  (a  +  b)  —  2abc. 

2.  2a?-2a2b  +  c3.  5.  5a2b-±ab2  +  21c. 

3.  2a2  +  3bc-5.  6.  7 a2  +  (a-  b)  (a-c). 

When  a  =  4,  b  =  2,  c  =  3,  and  d  =  l,  of 


a 


2  J,2 


1.15a-7(b+c)-d.         10.  ^  +  |  +  ^. 

8.  25a2-7(b2  +  c2)  +  d\     11.         4        + 1 . 

„«      £»      c  '25  a  —  30  c—  d 

9.  -  H 1-  -  .  12.  . 

bed  b  +  c 

When  a  =  \,  b  =  \,  c  =  \,  and  x  =  2,  of 

IS.  (2  a  +  3  b  +  5  c)  ($  a  +  3  b  -  5  e)  (2  a-  3  b  +  15  c). 


DEFINITIONS   AND   NOTATION.  H 

15.  x*  -  (2  a  +  3  b)  x3  +  (3  a  -  2  b)  x2-  ex  +  be. 
When  a=b,  and  b  =  £,  of 

16  5  a  +  ^  -  C3  a  -  (2  a  -  ^)] 

17  13  a  +  3  b  +  {7  Q  +  b)  +  [3  a  +  8  (4  a  -  b)~\) 

2a  +  Sb 

When  b  =  3,  c  =  4,  d  =  6,  and  e  =  2,  of 

18.  V 27T- v' 27+ y/2Z  19.  V  3^7+ ^"9^-^27. 

When  «  =  16,  &  =  10;  cc  =  5,  and  y=l,  of 

20.  (b  -  x)  (y/a  +  b)  +  ^  (a  -  b)  (x  +  y). 

48.     What  is  the  coefficient  of 

1.  x  in  3  n2  x  ?  3.  x  y  in  —  20  m?  xyzz? 

2.  a  cs  in  a  J2  c3  d4  ?  4.  ra2  w3  in  5  a8  m2  a;  ?i3  ? 

What  is  the  degree  of 

5.  3ax?        6.  2m*nx*?        7.  a2bsc2d5?        8.  2mcr2y3.?? 

Arrange  the  following  expressions  according  to  the  increas- 
ing powers  of  x : 

9.  2cc2-3a;  +  x3  +  l-4a;4. 

10.  3  x  y*  —  5  xs  y  +  yi  —  x4  —  x2  y2. 

Arrange  the  following  expressions  according  '  o  the  decreas- 
ing powers  of  a : 

11.  1- a2 -2  a  +  a3 +  2  a*. 

12.  aJ3-i4  +  a4-4«2i2-3a35. 


12  ALGEBRA. 


NEGATIVE   QUANTITIES. 

49.  The  signs  +  and  — ,  besides  indicating  the  operations 
of  addition  and  subtraction,  are  also  used,  in  Algebra,  to  indi- 
cate the  nature  or  quality  of  the  quantities  to  which  they  are 
prefixed. 

To  illustrate,  let  us  suppose  a  person,  having  a  property  of 
$  500,  to  lose  $  150,  then  gain  $  250,  and  finally  to  incur  a  debt 
of  $  450 ;  it  is  required  to  find  the  amount  of  his  property. 

Since  gains  have  an  additive  effect  on  property,  and  debts  or 
losses  a  subtractive  effect,  we  may  indicate  these  different 
qualities  algebraically  by  prefixing  the  signs  +  and  —  to  them, 
respectively ;  thus,  we  should  represent  the  transactions  as 
follows, 

$  500  -  $  150  +  $  250  -  $  450 ; 

which  reduces  to  $  150,  the  amount  required. 

But  suppose,  having  a  property  of  $  500,  he  incurs  a  debt 
of  $  700 ;  we  should  represent  the  transaction  algebraically  as 

follows, 

$500-1700; 

or,  as  incurring  a  debt  of  $  700  is  equivalent  to  incurring  two 
debts,  one  of  $  500  and  the  other  of  $ 200,  the  transaction  may 
be  expressed  thus, 

$  500  -  $  500  -  $  200. 

Now  since,  by  Art.  33,  $  500  and  —  $  500  neutralize  each 
other,  we  have  remaining  the  isolated  negative  quantity 
—  $200  as  the  algebraic  representative  of  the  required  prop- 
erty. In  Arithmetic,  we  should  say  that  he  owed  or  was  in 
debt  $200;  in  Algebra,  we  make  also  the  equivalent  state- 
ment that  his  property  amounts  to  —$200. 

In  this  way  we  can  conceive  the  possibility  of  the  indepen- 
dent existence  of  negative  quantities;  and  as,  in  Arithmetic, 
losses  may  be  added,  subtracted,  multiplied,  etc.,  precisely  as 
though  they  were  gains,   so,  in   Algebra,   negative  quantities 


DEFINITIONS   AND   NOTATION.  13 

may  be  added,  subtracted,  multiplied,  etc.,  precisely  as  though 
they  were  positive. 

The  distinction  of  positive  and  negative  quantities  is  applied 
in  a  great  many  cases  in  the  language  of  every-day  life  and  in 
the  mathematical  sciences.  Thus,  in  the  thermometer,  we 
speak  of  a  temperature  above  zero  as  +,  and  one  below  as  — ; 
for  instance,  +25°  means  25°  above  zero,  and  —10°  means 
10°  below  zero.  In  navigation,  north  latitude  is  considered 
-j-,  and  south  latitude  — ;  longitude  west  of  Greenwich  is  con- 
sidered +,  and  longitude  east  of  Greenwich  — ;  for  example, 
a  place  in  latitude  —  30°,  longitude  +  95°,  would  be  in  latitude 
30°  south  of  the  equator,  and  in  longitude  95°  west  of  Green- 
wich. And,  in  general,  when  we  have  to  consider  quantities 
the  exact  reverse  of  each  other  in  quality  or  condition,  we 
may  regard  quantities  of  either  quality  or  condition  as  posi- 
tive, and  those  of  the  opposite  quality  or  condition  as  negative. 
It  is  immaterial  which  quality  we  regard  as  positive  ;  but  hav- 
ing assumed  at  the  commencement  of  an  investigation  a  certain 
quality  as  positive,  we  must  retain  the  same  notation  through- 
out. 

The  absolute  value  of  a  quantity  is  the  number  represented 
by  that  quantity,  taken  independently  of  the  sign  affecting  it. 
Thus,  2  and  —  2  have  the  same  absolute  value. 

But  as  we  consider  a  person  who  owns  $  2  as  better  off 
than  one  who  owes  $2,  so,  in  Algebra,  we  consider  +  2  as 
greater  than  —  2 ;  and,  in  general,  any  positive  quantity, 
however  small,  is  considered  greater  titan  any  negative  quan- 
tity. 

Also,  as  we  consider  a  person  who  owes  $  2  as  better  off  than 
one  who  owes  $3,  so,  in  Algebra,  we  consider  — 2  as  greater 
than  —3;  and,  in  general,  oftivo  negative  quantities,  that  is 
regarded  as  the  greater  which  has  the  less  number  of  units,  or 
which  has  the  smaller  absolute  value. 

Again,  as  we  consider  a  person  who  has  no  property  or  debt 
as  better  off  than  one  who  is  in  debt,  so,  in  Algebra,  zero  is 
considered  greater  than  any  negative  quantity. 


14  ALGEBRA. 


II.  — ADDITION. 

50.  Addition,  in  Algebra,  is  the  process  of  collecting  two 
or  more  quantities  into  one  equivalent  expression,  called,  the 
sum. 

51.  In  Arithmetic,  when  a  person  incurs  a  debt  of  a  certain 
amount,  we  regard  his  property  as  diminished  by  the  amount 
of  the  debt.  So,  in  Algebra,  using  the  interpretation  of  nega- 
tive quantities  as  given  in  Art.  49,  adding  a  negative  quantity 
is  equivalent  to  subtracting  an  equal  positive  quantity.  Thus, 
the  sum  of  a  and  —  b  is  obtained  by  subtracting  b  from  a,  giv- 
ing as  a  result  a  —  b. 

Hence,  the  addition  of  monomials  is  indicated  by  uniting 
the  quantities  with  their  respective  signs.  Thus,  the  sum  of 
a,  —  b,  c,  d,  —  e,  and  — f,  is 

a—b+c+d—e  —f. 

The  addition  of  polynomials  is  indicated  by  enclosing  them 
in  parentheses  (Art.  25),  and  uniting  the  results  with  +  signs. 
Thus,  the  sum  of  a  +  b  and  c  —  d  is 

(a  +  b)  +  (c  —  d). 

52.  Let  it  be  required  to  add  c  —  d  to  a  +  b. 

If  we  add  c  to  a  +  b,  the  sum  will  be  a  +  b  +  c.  But  we 
have  to  add  to  a  +  b  a  quantity  which  is  d  less  than  c.  Conse- 
quently our  result  is  d  too  large.  Hence  the  required  sum  will 
be  a  +  b  +  c  diminished  by  d,  or  a  +  b  +  c  —  d. 

Hence,  the  addition  of  polynomials  may  also  be  indicated  by 
uniting  their  terms  with  their  respective  signs. 

53.  Let  it  be  required  to  add  2  a  and  3  a. 
By  Art.  32,  2  a  =  a  +  a, 

and  3  a  =  a  +  a  +  a. 


ADDITION.  15 

Hence  (Art.  52)  the  sum  of  2  a  and  3  a  is  indicated  by 

a  -\-  a  +  a  +  a  +  a, 
which,  by  Art.  32,  is  equal  to  5  a.     Hence,  2a  +  3a  =  5a. 

54.  Let  it  be  required  to  add  —  3  a  and  —  2  a. 
By  Art.  32,  —  3  a  =  —  a  —  a  —  a, 

and  —  2  a  =  —  a  —  a. 

Hence  (Art.  52),  the  sum  of  —  3  a  and  —  2  a  is  indicated  by 

—  a  —  a  —  a  —  a  —  a, 

or  —  5  a  (Art.  32).     Hence,  —  3  a  —  2  a  =  —  5  a. 

From  our  ideas  of  negative  quantities  (Art.  49),  we  may  ex- 
plain this  result  arithmetically  as  follows  : 

If  a  person  has  two  debts,  one  of  $  3  and  the  other  of  $  2, 
he  may  be  considered  to  be  in  debt  to  the  amount  of  $  5. 

55.  Let  it  be  required  to  add  4  a  and  —  2  a. 

4  a  =  a  +  a+  a  +  a, 

and  —  2«  =  —  a  —  a. 

Hence,  the  sum  of  4  a  and  —  2  a  is  indicated  by 

a  +  a  +  a  +  a  —  a  —  a. 

Now,  by  Art.  33,  the  third  and  fourth  terms  are  neutralized 
by  the  fifth  and  sixth,  leaving  as  the  result  a  +  a,  or  2  a. 
Hence,  4  «  —  2  a  =  2  a. 

We  may  explain  this  result  arithmetically  as  follows : 
If  a  person  has  $4  in  money,  and  incurs  a  debt  of  $2,  his 
property  may  be  considered  to  amount  to  $  2. 

56.  Let  it  be  required  to  add  —  4  a  and  2  a. 

—  4:a  =  —  a  —  a  —  a  —  a, 
and  2a  =  ffl+«. 


16  ALGEBRA. 

Hence,  the  sum  of  —  4  a  and  2  a  is  indicated  by 

—  a  —  a  —  a  —  a  +  a  +  a. 

The  third  and  fourth  terms  neutralize  the  fifth  and  sixth, 
leaving  as  the  result  —  a  —  a  or  —2a.     Hence, 

—  4ta  +  2a  =  —  2a. 

We  may  explain  this  result  arithmetically  as  follows  : 
If  a  person  has  $  2  in  money,  and  incurs  a  debt  of  84,  he 
may  be  considered  to  be  in  debt  to  the  amount  of  $2. 

.57.  From  Arts.  55  and  56  we  derive  the  following  rule 
for  the  addition  of  two  similar  (Art.  34)  terms  of  opposite 
sign: 

To  add  two  similar  terms,  the  one  positive  and  the  other 
negative,  subtract  the  smaller  coefficient  from  the  larger,  affix 
to  the  result  the  common  symbols,  and  prefix  the  sign  of  the 
larger. 

For  example,  the  sum  of  7  x  y  and  —  3xy  is  4:xy, 

the  sum  of  3  a2  bs  and  —  11  a~  b3  is  —  8  a2  b3. 

58.  In  Arithmetic,  when  adding  several  quantities,  it 
makes  no  difference  in  which  order  we  add  them ;  thus, 
3  +  5  +  9,  5  +  3  +  9,  9+3  +  5,  etc.,  all  give  the  same  result, 
17.  So  also  in  Algebra,  it  is  immaterial  in  what  order  the 
terms  are  united,  provided  each  has  its  proper  sign.     Thus, 

—  b  +  a  is  the  same  as  a  —  b. 

Hence,  in  adding  together  any  number  of  similar  terms, 
some  positive  and  some  negative,  we  may  add  the  positive 
terms  first,  and  then  the  negative,  and  finally  combine  these 
two  results  by  the  rule  of  Art.  57. 

Thus,  in  finding  the  sum  of  2  a,  —  a,  la,  6  a,  —  4  a,  and 

—  5  a,  the  sum  of  the  positive  terms  2  a,  7  a,  and  6  a,  is  15  a, 
and  the  sum  of  the  negative  terms  —  a,  —  4  a,  and  —  5  a, 
is  —  10  a  ;  and  the  sum  of  15  a  and  —  10  a  is  5  a. 

59.  Let  it  be  required  to  add  6  a  —  7  x,  3  x  —  2  a  +  3  y, 
and  2  x  —  a  —  mn. 


ADDITION.  1 7 

We  might  obtain  the  sum  in  accordance  with  Art.  52,  by- 
uniting  the  terms  by  their  respective  signs,  and  combining 
similar  terms  by  the  methods  previously  given.  It  is  however 
customary  in  practice,  and  more  convenient,  to  set  the  expres- 
sions down  one  underneath  the  other,  similar  terms  being  in 
the  same  vertical  column  ;  thus, 

6  a  —  1  x 
—  2  a  +  3 x  +  3  y 
—  a  +  2  x  —mn 


3  a  —  2  x  +  3  y  —  m  n. 

It  should  be  remembered  that  only  similar  terms  can  be 
combined  by  addition  ;  and  that  the  algebraic  sum  of  dissimilar 
terms  can  only  be  indicated  by  uniting  them  by  their  respective 
signs. 

60.  From  the  preceding  principles  and  illustrations  is  de- 
rived the  following 

RULE. 

To  add  together  two  or  more  expressions,  set  them  down  one 
underneath  the  other,  similar  terms  being  in  the  same  vertical 
column.  Find  the  sum  of  the  similar  terms,  and  to  the  result 
obtained  unite  the  dissimilar  terms,  if  any,  by  their  respective 
signs. 


EXAMPLES. 

1. 

2. 

3. 

4. 

5. 

la 

—  6  m 

13  n 

—  4  a  x 

2a2b 

3  a 

m 

n 

—  3  ax 

-a-b 

a 

— 11  m 

—  20n 

a  x 

11  a-  b 

5  a 

—  5m 

6  n 

—  lax 

-5a2b 

11a 

—  m 

8n 

—  ax 

±a2b 

a 

20  m 

—  n 

12  ax 

-9aH 

18  ALGEBRA. 

6.  7.  8. 

la —       mp2  2  a  — 3  x  ab  +     c  d 

a  +    6  mp2  —  a  +  4x  —  ab  -\-     cd 

— 11  a  —   3  mp2  a  +     x  3  a  b  —  2  cd 

8  a  + 11  m  j92  5  a  —  7  a;  lab  —  5  cd 

—  9  a—   1  mp2  —4:  a—    x  —4tab  +  6cd 

18  a  —  15  mp2  —3  a  +  1  x  2  ab  —  5cd 


Find  the  sum  of  the  following : 

9.   4,xy  z,  —  3xy  z,  —  5xy  z,  6x  y  z,  —  9xy  z,  and  3 x y  z. 

10.  5  m  n2  —  8x2y,  —  m  n2  +  x2  y,  —  6m  n2  —  3x2y,  4:mn2 
+  1  x2y,  2  m  n2  +  3  x2  y,  and  —  ra  ri2  —  2x2y. 

11.  3a2  +  2ab  +  4,b2,    5a2-Sab  +  b2,    -a2+5ab-b2, 
18  a2 -20  ab-  19  b2,  and  14  a2  -  3  a  b  +  20  b2. 

12.  2a  —  5b  —  c  +1,  3b  —  2-6a  +  8c,  c  +  3a-4,  and 
1  +  2  b  -  5  c. 

13.  6x  —  3y+lm,    2  n  —  x  +  y,    2  y  —  4x—  5  m,    and 
m  +  n  —  y. 

14.  2  a  -  3  b  +  4  <7,    2  &  -  3  d  +  4  c,    2  cZ  -  3  e  +  4  a,    and 
2c-3a  +  4i. 

15.  3  cc  —  2  y  —  z,  3  y  —  5  x  —  1  z,  8  z  —  y  —  x,  and  4  x. 

16.  2  m  —  3n  +  5r  —  t,  2  n  —  6  t  —  3  r  —  m,   r  +  3  m  —  5n 
+  4t,  and  3  t  —  2  r  +  1  n  —  4  m. 

17.  4:inn  +  3  ab  —  4  c,  3  x  —  4  a  b  +  2  m  n,  and  3  m2  —  4  p. 

18.  3  a  +  b  —  10,    c  —  d  —  a,    —4c  +  2a  —  3b  —  l,    and 
4  a;2  +  5  -  18  m. 

19.  4a;8_5a8_  5ax2+6a2x,  6as  +  3x*  +  Aax2+  2a2x, 
-11  Xs +19  ax2-  15  a2  x,  and  10  x3  +  1  a2  x  +  5  a3  -  18  a  x2. 

20.  la  —  5if,  S^x  +  2a,  oif  —  \/x,  and  —  9a  +  7tfx. 

21.  3  a  b  +  3  («  +  b),    -  a  b  +  2  (a  +  b),    7  a  b  —  4  (a  +  b), 
and  —  2  a  b  +  6  (a  +  b). 

22.  lsjy-4(a-b),  6  \J  y  +  2  (a  -  b),  2  ^  y  +  (a-b),  and 
sjy  —  3(a  —  b). 


SUBTRACTION.  19 


III.  — SUBTRACTION. 

61.  Subtraction,  in  Algebra,  is  the  process  of  finding  one 
of  two  quantities,  when  their  sum  and  the  other  quantity  are 
given. 

Hence,  Subtraction  is  the  converse  of  Addition. 

The  Minuend  is  the  sum  of  the  quantities. 

The  Subtrahend  is  the  given  quantity. 

The  Remainder  is  the  required  quantity. 

As  the  remainder  is  the  difference  between  the  minuend 
and  subtrahend,  subtraction  may  also  be  defined  as  the  process 
of  finding  the  difference  between  two  quantities. 

62.  Subtraction  may  be  indicated  by  writing  the  subtra- 
hend after  the  minuend,  with  a  —  sign  between  them.  Thus, 
the  subtraction  of  b  from  a  is  indicated  by 

a  —  b. 

In  indicating  subtraction  in  this  way,  the  subtrahend,  if  a 
negative  quantity  or  a  polynomial,  should  be  enclosed  in  a 
parenthesis.  Thus,  the  subtraction  of  —b  from  a  is  indi- 
cated by 

and  the  subtraction  of  b  —  c  from  a  by 

a—  (b  —  c). 

63.  Let  it  be  required  to  subtract  b  —  c  from  a. 
According  to  the   definition  of  Art.  61,  we  are  to  find  a 

quantity  which  when  added  to  b  —  c  will  produce  a  ;  this 
quantity  is  evidently  a  —  b  +  c,  which  is  the  remainder  re- 
quired. 

Now,  if  we  had  changed  the  sign  of  each  term  of  the  sub- 
trahend, giving  —  b  +  c,  and  had  added  the  resulting  expres- 
sion to  a,  we  should  have  arrived  at  the  same  result,  a  —  b  +  c. 


20  ALGEBRA. 

Hence,  to  subtract  one  quantity  from  another,  we  may  change 
the  sign  of  each  term  of  the  subtrahend,  and  add  the  residt  to 
the  minuend. 

64.  1.    Let  it  be  required  to  subtract  3  a  from  8  a. 
According  to  Art.  63,  the  result  may  be  obtained  by  adding 

—  3  a  to  8  a,  giving  5  a  (Art.  55). 

2.  Subtract  8  a  from  3  a. 

By  Art.  63,  the  result  is  3  a  —  S  a  or  —5a  (Art.  56). 

3.  Subtract  —  2  a  from  3  a. 
Result,  3  a  +  2  a  or  5  a. 

4.  Subtract  3  a  from  —2  a. 

Result,  —  2a-3«or  —5  a. 

5.  Subtract  — 2  a  from  —5  a. 
Result,  —  5  a  +  2  a  or  —3  a. 

6.  Subtract  —5a  from  —2  a. 
Result,  —  2  a  +  5  a  or  3  a. 

65.  In  Arithmetic,  addition  always  implies  augmentation, 
and  subtraction  diminution.  In  Algebra  this  is  not  always 
the  case  ;  for  example,  in  adding  —  2  a  to  5  a  the  sum  is  3  a, 
which  is  smaller  than  5  a  ;  also,  in  subtracting  —2  a  from  5  a 
the  remainder  is  7  a,  which  is  larger  than  5  a.  Thus,  the 
terms  Addition,  Subtraction,  Sum,  and  Remainder  have  a 
much  more  general  signification  in  Algebra  than  in  Arith- 
metic. 

66.  From  Art.  63  we  derive  the  following 

RULE. 

To  subtract  one  expression  from  another,  set  the  subtrahend 
underneath  the  minuend,  similar  terms  being  in  the  same  ver- 
tical column.  Change  the  sign  of  each  term  of  the  subtrahend 
from  +  to  — ,  or  from  —  to  + ,  and  add  the  restdt  to  the 
minuend. 


USE   OF   PAKENTHESES.  21 


EXAMPLES. 

1. 

2. 

3. 

4. 

5. 

27  a 

17  x 

-13  3, 

- 10  m  n 

5  a2  b 

13  a 

-11b 

Ay          - 

- 18  m  n 

UaH 

ab  +     cd  —     ax  7x+5y—3a 

Aab  —  3  cd-\-  Aax  x—7y+ha—A 

8.  9. 

7  abc-llx  +  5y-   48  5  \Ja-  3  y2  +  7  a  -6 

llabc+    3x  +  7?/  +  100  3<Ja+     y2-5a-7 

10.  Subtract  —5b  from  -  12  J. 

11.  From  31  x2  -  3  y2  +  a  b  take  17  x2  +  5  if  -  4  a  b  +  7. 

12.  Subtract  a  —  b  +  c  from  a  +  b  —  c. 

13.  Subtract  6a  —  3^  —  5c  from  Qa-\-3b  —  5  c  +  1. 

14.  From  3m-5ft+r-2s  take  2  r  +  3  n  —  m  —  5  s. 

15.  Take  4  a  —  b  +  2  i  —  5  d  from  a"  —  3  &  +  a  -  c. 

16.  From  m2  +  3  nz  take  —  4  m'2  —  6  ?i8  +  71  cc. 

17.  From  a  +  b  take  2«-25  and  —  a  +  b. 

18.  From  a  — b  —  c  take  —  a  +  b  +  c  and  a  —  b  +  c. 


IV.  — USE  OF  PARENTHESES. 

67.  The  use  of  parentheses  is  very  frequent  in  Algebra, 
and  it  is  necessary  to  have  rules  for  their  removal  or  introduc- 
tion 


22  ALGEBRA. 

68.  Let  it  be  required  to  indicate  the  addition  of  3  a.  and 
5  b  —  c  +  2  d ;  this  we  may  do  by  placing  the  latter  expression 
in  a  parenthesis,  prefixing  a  +  sign,  and  writing  after  the 
former  quantity,  thus  : 

3  a  +  (5  b  -  c  +  2  d). 

If  the  operation  be  performed,  we  obtain  (Art.  60), 

3a  +  5b  —  c  +  2d. 

69.  Again,  let  it  be  required  to  indicate  the  subtraction  of 
5b  —  c-\-2d  from  3  a  ;  this  we  may  do  by  placing  the  former 
expression  in  a  parenthesis,  prefixing  a  —  sign,  and  writing 
after  the  latter  quantity,  thus  : 

3a-(5b-c  +  2d). 

If  the  operation  be  performed,  we  obtain  (Art.  66), 

3  a  —  5  b  +  c  —  2d. 

70.  It  will  be  observed  that  in  the  former  case  the  signs 
»f  the  terms  within  the  parenthesis  are  unchanged  when  the 
parenthesis  is  removed ;  while  in  the  latter  case  the  sign  of 
each  term  within  is  changed,  from  +  to  — ,  or  from  —  to  +. 
Hence,  we  have  the  following  rule  for  the  removal  of  a  paren- 
thesis : 

If  the  parenthesis  is  preceded  by  a  -f-  sign,  it  may  be  re- 
moved if  the  sign  of  every  enclosed  term  be  unchanged;  and 
if  the  parenthes-is  is  preceded  by  a  —  sign,  it  may  be  removed 
if  the  sign  of  every  enclosed  term  be  changed. 

71.  To  enclose  any  number  of  terms  in  a  parenthesis,  we 
take  the  reverse  of  the  preceding  rule : 

Any  number  of  terms  may  be  enclosed  in  a  parenthesis,  with 
a  +  sign  prefixed,  if  the  sign  of  every  term  enclosed  be  un- 
changed;  and  in  a  parenthesis,  with  a  —  sig?i  prefixed,  if 
the  sign  of  every  term  enclosed  be  changed. 


USE  OF   PARENTHESES.  23 

72.  As  the  bracket,  brace,  and  vinculum  (Art.  25)  have  the 
same  signification  as  the  parenthesis,  the  rules  for  their  re- 
moval or  introduction  are  the  same.  It  should  be  observed 
in  the  case  of  the  vinculum,  that  the  sign  apparently  prefixed 
to  the  first  term  underneath  is  in  reality  the  sign  of  the  vin- 
culum;  thus,  +  a  —  b  signifies  +  (a  —  V),  and  —  a  —  b  signi- 
fies —  (a  —  b). 

73.  Parentheses  will  often  be  found  enclosing  others ;  in 
this  case  they  may  be  removed  successively,  by  the  preceding 
rule ;  and  it  is  better  to  begin  by  removing  the  inside  pair. 

74.  1.    Remove   the   parentheses   from   3  a  —  (2  a  —  5)  — 

(-a +  7). 

Result,  3a  —  2a  +  5  +  a  —  7  —  2a  —  2. 

2.    Remove  the  parentheses  etc.,  from 


6  a  -  [3  a  +  (2  a  -  { 5  a  -  [4  a  -  a  -  2] } )]. 

In  accordance  with  Art.  73,  we  remove  the  vinculum  first, 
and  the  others  in  succession.     Thus, 


6  a  —  [3  a  +(2  a -{5  a -[4  a- a- 2]})] 
=  6  a  -  [3  a  +  (2  a  -  {5  a  -  [4  a  -  a  +  2]})] 
=  6  a  —  [3  a  +  (2  a—  {5  a  —  4  a  +  a  —  2})] 

=  6  a  —  [3  a  +  (2  a  —  5  a  +  4  a  —  a  +  2)] 

=  Qa  —  [3a  +  2a  —  5a  +  4a  —  a  +  2^\ 

=  6a  —  3a  —  2a  +  5a  —  4:a  +  a  —  2  =  3a  —  2,  Ans. 

3.   Enclose  the  last  three  terms  of  a  —  b~ c  +  d+  e  —f  in 
a  parenthesis  with  a  —  sign  prefixed. 

Result,  a  —  b  —  c—(—d  —  e+f). 


24  ALGEBRA. 

EXAMPLES. 

Remove  the  parentheses,  etc.,  from  the  following  : 

4.  a  —  (b  —  c)  +  (d  —  e). 

5.  3a-(2a-{a  +  2}). 

6.  5  x  —  (2  x  —  3  y)  —  (2  x  +.  4  y). 

7.  a  —  b  +  c  —  {a  +  b  —  c)  —  {c  +  b  —  a). 

8.  in'  —  2n+  (a  —  n  +  3  ?m2)  —  (5  a  +  3  w  —  m2). 

9.  2  m  -  [n  —  {3  wi -(2w-  m) } ]. 


10.   8x  —  (5x  —  [4  a;  —  ?/  —  &■])  —  (—  a;  —  3  y). 


11.  2«-[5J+  {3c-(a+[2ft-3a  +  4c])}]. 

12.  3c+(2a-[5c-{3a  +  c-4a}-]). 


13.    6  a  -  [5  a  -  (4  a  -  { -  3  a  -  [2  a- a-  1]})]. 


14.   2  m  -  [3  m  -  (5  m - 2)  - { m -  (2  wi -3m  +  4)}]. 

75.     As  another  application  of  the  rule  of  Art.  70,  we  have 
the  following  four  results : 

+  (+  a)  is  equivalent  to  +  a ; 
+  ( —  a)  is  equivalent  to  —  a  ; 

—  (+  a)  is  equivalent  to  —  a  ; 

—  (—a)  is  equivalent  to  +  a. 


V.  —  MULTIPLICATION. 

76.  Multiplication,  in  Algebra,  is  the  process  of  taking 
one  quantity  as  many  times  as  there  are  units  in  another 
quantity. 

The  Multiplicand  is  the  quantity  to  be  multiplied  or  taken. 

The  Multiplier  is  the  quantity  by  which  we  multiply. 

The  Product  is  the  result  of  the  operation. 

The  multiplicand  and  multiplier  are  often  called  factors. 


MULTIPLICATION.  25 

77.  The  product  of  the  factors  is  the  same,  in  whatever 
order  they  are  taken. 

For  we  know,  from  Arithmetic,  that  the  product  of  two 
numbers  is  the  same,  in  whatever  order  they  are  taken  ;  thus 
we  have  3  X  4  or  4  X  3  eacli  equal  to  12.  Similarly,  in  Alge- 
bra, where  the  symbols  represent  numbers,  we  have  a  X  b  or 
b  X  a  each  equal  to  a  b  (Art.  14). 

78.  Let  it  be  required  to  multiply  a  —  b  by  c. 

By  Art.  77,  multiplying  a  —  b  by  c  is  the  same  as  multiply- 
ing c  by  a  —  b.  To  multiply  c  by  a  —  &,  we  multiply  it  first 
by  a,  and  then  by  b,  and  subtract  the  second  result  from  the 
first,  e  multiplied  by  a  gives  a  c,  and  multiplied  by  b  gives 
b  c.     Subtracting  the  second  result  from  the  first  we  have 

a c  —  b  c 
the  product  required. 

79.  Let  it  be  required  to  multiply  a  —  b  by  c  —  d. 

To  multiply  a  —  b  by  c  —  d,  we  multiply  it  first  by  c,  and 
then  by  d,  and  subtract  the  second  result  from  the  first.  By 
Art.  78,  a  —  b  multiplied  by  c  gives  ac  —  bc,  and  multiplied 
by  d  gives  ad  —  bd.  Subtracting  the  second  result  from  the 
first,  we  have 

ac  —  bc  —  ad+bd 

the  product  required. 

80.  We  observe  in  the  result  of  Art.  79, 

1.  The  product  of  the  positive  term  a  by  the  positive  term 
c  gives  the  positive  term  a  c. 

2.  The  product  of  the  negative  term  —b  by  the  positive 
term  c  gives  the  negative  term  —be. 

3.  The  product  of  the  positive  terra  a  by  the  negative  term 
—  d  gives  the  negative  term  —  a  d. 

4.  The  product  of  the  negative  tern>  —b  by  the  negative 
term  —  d  gives  the  positive  term  b  d. 


26  ALGEBRA. 

From  these  considerations  we  can  state  what  is  known  as 
the  Rule  of  Signs  in  Multiplication,  as  follows: 

+  multiplied  by  +,  and  —  multiplied  by  — ,  produce  +  ; 
+  multiplied  by  — ,  and  —  multiplied  by  +  ,  produce  — . 

Or,  as  may  he  enunciated  for  the  sake  of  hrevity  with  regard 
to  the  product  of  any  two  terms, 

Like  signs  produce  +  ,  and  unlike  signs  produce  — . 

81.  Let  it  he  required  to  multiply  7  a  by  2  b. 

Since  (Art.  77)  the  factors  may  he  written  in  any  order,  we 
have  7ax2b  =  7x2xaXb  =  14:ab.     Hence, 

The  coefficient  of  the  product  is  equal  to  the  product  of  the 
coefficients  of  the  factors. 

82.  Let  it  he  required  to  multiply  a3  hy  a2. 

By  Art.  17,  as  means  sX«X«}  and  a2  means  aXa;  hence, 

a!Xfl2  =  «X»X«X«X(i=«5.     Hence, 

The  exponent  of  a  letter  in  the  product  is  equal  to  the  sum 
of  its  exponents  in  the  factors.  ' 

Or,  in  general,  am  X  an  =  am  +  n. 

83.  In  Multiplication  we  may  distinguish  three  cases. 

CASE    I.         • 

84.  WJien  both  factors  are  monomials. 

From  Arts.  80,  81,  and  82  is  derived  the  following  rule  for 
the  product  of  any  two  monomials. 

RULE. 

Multiply  the  numerical  coefficients  together  ;  annex  to  the 
residt  the  letters  of  both  monomials,  giving  to  curb  letter  an 
exponent  equal  to  the  sum  of  its  exponents  in  the  factors.  Make 
the  product  +  when  the  two  factors  have  the  same  sign,  and  — 
when  they  have  different  signs. 


MULTIPLICATION.  27 

EXAMPLES. 

1.  Multiply  2  «"  by  3  a2. 

2«4x3a2  =  6a6,  Ans. 

2.  Multiply  a3  b2  c  by  -  5  a2  b  d. 

a8  b2  c  X  —  5  a2  b  d  —  —  5  a5  bs  c  d,  Ans. 

3.  Multiply  —  7  xm  by  —  5  sen. 

—  7  a;m  x  —  5  xn  =  35  a:m+n,  ^4?zs. 

4.  Multiply  3  a  (a;  —  y)2  by  4«3(x-  ?/). 

3  a  (a  -  y)2  X  4  a3  (x-y)  =  12  a4  (a;  -  y)3,  Ans. 

Multiply  the  following  : 

5.  15  m5  w6  by  3  m  n.        12.  —  12  a2  x  by  —  2  a2  y. 

6.  3  a  6  by  2  a  e.  13.  3  am  6n  by  —  5  an  br. 

7.  17  a  b  c  by  —  8  a  b  c.    14.  —  4  xm  ?/"  by  —  xn  yn  zb. 
8.-17  a4  c2  by  -  3  a2  c2.  15.  2  am  5"  by  5  a3  b. 

9.   11  n2  y  by  —  5  w6 «.      16.   —  7  mn  x2  by  mn  #r  y2. 

10.  4a6by3aiy2.  17.    2  m  (a  -  b)2  by  m  (a  -  b). 

11.  —  6  a  b2  c  by  a3  b  m.     18.    7  a  (x  —  y)  hy  —3  a2  b  (x  —  y). 

19.  Find  the  continued  product  of  8  a  x2,  2  a3  y,  and  4  a;3  v/4. 

20.  Find   the   continued    product    of  2  a  c2,    —  4  a  c3,    and 

-3«J2. 

CASE    II. 

85.      Wlien  one  of  the  factors  is  a  polynomial. 
From  Art.  78  we  have  the  following 

RULE. 

Multiply  each  term  of  the  multiplicand  by  the  multiplier, 

remembering  that  like  signs  produce  +,  and  unlike  signs  pro- 
duce — . 


28  ALGEBRA. 

EXAMPLES. 

1.    Multiply  3  x  —  y  by  2  x  y. 

3  x  —  y 
2  x  y 


6  x'2  y  —  2  x  y2,  Ans. 
2.    Multiply  3  a  —  5  x  by  —  4  ra. 

3  a  —  5  x 
—  4  m 


,3 


.2 


— 12  a  m  +  20  ra  #,  ^4?is. 
Multiply  tbe  following : 

3.  x2-2z-3by  4z.  7. -x4-10a;3  +  5by-2x- 

4.  8  a2  6  c  -  fZ  by  5  a  f/2.  8.  a2  +  13  a&  -  6  62  by  4  a  b2 

5.  3  x2  +  6  x  —  7  by  —  2  xs.  9.  ra2  +  m  n  +  ?r  by  m  n. 

6.  3  ra2  —  5  ra  ?i  —  ?i2  by  —  2  m.    10.  5  —  6  a  —  8  a3  by  —  6  <x 

11.  5a;3-4x2-3z-2by-6r\ 

12.  «3  -  3  a2  b  +  3  a  b2  -  b*  by  a2  b2. 


CASE    III. 

86.      When  both  of  the  factors  are  polynomials. 

In  Art.  79  we  sbowed  that  tbe  product  of  a  —  b  and  c  —  d 
rnigbt  be  obtained  by  multiplying  a  —  b  by  c,  and  then  by  d, 
and  subtracting  tbe  second  result  from  tbe  first.  It  would 
evidently  be  equally  correct  to  multiply  a  —  b  by  c,  and  then 
by  —  d,  and  add  the  second  result  to  the  first.  On  this  we 
base  the  following  rule  for  finding  the  product  of  two  poly- 
nomials. 

RULE. 

Multiply  each  term  of  the  multiplicand  by  each  term  of  the 

multiplier,  remembering  that  like  signs  'produce  +,  and  unlike 
signs  produce  —,  and  add  the  partial,  products. 


MULTIPLICATION.  29 


EXAMPLES. 

1.    Multiply  3  a  -  2  b  by  2  a- 5b. 

3a  -2b 
2a  —5b 


6  a2—    Aab 

-  15  a  b  +  10  b2 

6  a2  — 19  a  b  +  10  6'2,  ^ws. 

The  reason  for  shifting  the  second  partial  product  one  place 
to  the  right,  is  that  in  general  it  enables  us  to  place  like  terms 
in  the  same  vertical  column,  where  they  are  more  conveniently 
added. 

2.    Multiply  x2  +  1  —  x3  —  x  by  x  +  1. 

1  —  X  +  X2  —  X3 

1  +  x 


1  —  X  +  X2  —  X3 

+  X  —  X2  +  X3  —  X* 

1  —x4,  Ans. 

It  is  convenient,  though  not  essential,  to  have  both  multi- 
plicand and  multiplier  arranged  in  the  same  order  of  powers 
(Art.  41),  and  to  write  the  product  in  the  same  order. 

Multiply  the  following : 

3.  3  x2  —  2  x  y  —  y2  by  2  x  —  Ay. 

4.  x2  +  2x  +  lhyx2-2x  +  3. 

5.  a  +  b  —  c  by  a  —  b  +  c. 

6.  3a-2bhy-2a  +  4.b. 

7.  a2  +  b2  +  ab  by  b  —  a. 

8.  1  +  x  +  x3  +  x2  by  a  x  —  a. 

9.  5  a2  -  3  a  b  +  4  b2  by  6  a  -  5  b. 
10.  3  x2  —  7  x  +  4  by  2  x2  +  9  x  —  5. 


30  ALGEBRA. 

11.  6  x  -  2  x-  -  5  -  a-3  by  x2  +  10  -  2  x. 

12.  2a3+5a>2-8a:-7by4-5:*;-3a:2. 

13.  a3  b  -  a2  b2  -  4  a  68  by  2  a2  5  -  a  b2. 

14.  xm  +  2ij  —  3xyn~l  by  4  *"•  +  5  y2  —  4  a4  yn. 

15.  6  a:4  -  3  x3- a:2 +6  a; -2  by  2x2  +  x  +  2. 

16.  m4  —  m3  w  +  rn2  n2  —  m  w3  +  %4  by  m  +  n. 

17.  ft3-3«2H3ai2-  63  by  a2-2ab  +  b\ 

87.  It  is  sometimes  sufficient  to  indicate  the  product  of 
polynomials,  by  enclosing  each  of  the  given  factors  in  a  paren- 
thesis, arid  writing  them  one  after  the  other,  with  or  without 
the  sign  X  between  the  parentheses.  When  the  indicated 
multiplication  is  performed,  the  expression  is  said  to  be  ex- 
panded or  developed. 

1.  Indicate  the  product  of  2  x2  —  3  x  y  +  6  by  3  x2  +  3  x  y  —  o. 
Kesult,  (2x2-3xy  +  6)  (3x2+3xy-5). 

EXAMPLES. 

2.  Expand  (3  a  +  4  b)  (2  a  +  b). 

3.  Expand  (a*  —  a3  x  +  a2  x2  —  a  x3  +  x4)  (a  +  x). 

4.  Develop  (a*  —  a:4)  X  («4  —  a,'4). 

5.  Develop  (am  —  a")  (2  a  —  an). 

6.  Expand  (1  +  x)  (1  +  a-4)  (1  —  a-  +  .r2  -  a-3). 

7.  Find  the  value  of  («  +  2  a)  («  —  3  x)  (a  +  4  x). 

8.  Expand  [«  (a2  —  3  a  +  3)  -  1]  x  [a  (a  -  2)  +  1]. 

88.  From  the  definition  of  Art.  76,  0  X  a  means  0  taken 
a  times.  Since  0  taken  any  number  of  times  produces  0,  it 
follows  that  0  x  a  =  0.     That  is, 

If  zero  be  multiplied  by  any  quantity}  the  product  is  equal 
to  zero. 


DIVISION.  31 

89.  Since  (+  a)  X  (+  b)  —ab,  and  (—  a)  X  (— &)  =ab, 
it  follows  that  in  the  indicated  product  of  two  factors,  all  the 
signs  of  both  factors  may  be  changed  without  altering  the  value 
of  the  expression.     Thus, 

(x  —  y)  {a  —  b)  is  equal  to  {y  —  x)  (J>  —  a). 

Similarly  we  may  show  that  in  the  indicated  product  of  any 
number  of  factors,  any  even  number  of  factors  may  In/re  their 
signs  changed  without  altering  the  value  of  the  expression. 

Thus,    (x  —  y)  (c  —  d)  (e  —f)  (g  —  h)  is  equal  to 
{y-x)  (c-d)  (f-e)  (g-  h),  or  to 
(y  —  x)  (d  —  c)  {f-e)  (h  —  g),  etc.;  but  is  not  equal  to 
(y-x){d~c)(f-e)(g-h). 


VI.  — DIVISION. 

90.  Division,  in  Algebra,  is  the  process  of  finding  one  of 
two  factors,  when  their  product  and  the  other  factor  are  given. 

Hence,  Division  is  the  converse  of  Multiplication. 

The  Dividend  is  the  product  of  the  two  factors. 
The  Divisor  is  the  given  factor. 
The  Quotient  is  the  required  factor. 

91.  Since  the  quotient  multiplied  by  the  divisor  produces 
the  dividend,  it  follows,  from  Art.  80,  that  if  the  divisor  and 
quotient  have  the  same  sign,  the  dividend  is  +  ;  and  if  they 
have  different  signs,  the  dividend  is  —  .     Hence, 

+  divided  by  +,  and  —  divided  by  — ,  produce  +  ; 
+  divided  by  — ,  and  —  divided  by  +,  produce  — . 
Hence,  in  division  as  in  multiplication, 

Like  signs  produce  +,  and  unlike  signs  produce  — . 


32  ALGEBRA. 

92.  Let  it  be  required  to  find  the  quotient  of  14  a  b  divided 
by  7  a. 

Since  the  quotient  is  such  a  quantity  as  when  multiplied  by 
the  divisor  produces  the  dividend,  the  quotient  required  must 
be  such  a  quantity  as  when  multiplied  by  7  a  will  produce 
14  a  b.     That  quantity  is  evidently  2  b.     Hence, 

The  coefficient  of  the  quotient  is  equal  to  the  coefficient  of 
the  dividend  divided  by  the  coefficient  of  the  divisor. 

93.  Let  it  be  required  to  find  the  quotient  of  a5  divided 
by  a3. 

The  quotient  required  must  be  such  a  quantity  as  when 
multiplied  by  a3  will  produce  a5.  That  quantity  is  evidently 
a2.     Hence, 

The  exponent  of  a  letter  in  the  quotient  is  equal  to  its  expo- 
nent in  the  dividend  diminished  by  its  exponent  in  the  divisor. 

Or,  in  general,  am  -f-  an  =  am~n. 

94.  If  we  apply  the  rule  of  Art.  93  to  finding  the  quotient 
of  a™  divided  by  am,  we  have  am  -f-  am  =  am~m  =  a0. 

Now,  according  to  the  previously  given  definition  of  an  ex- 
ponent (Art.  17),  a°  has  no  meaning,  and  we  are  therefore  at 
liberty  to  give  to  it  any  definition  we  please.  As  am  —  am  =  1, 
we  should  naturally  define  a°  as  being  equal  to  1 ;  and  as  a 
may  represent  any  quantity  whatever, 

Any  quantity  whose  exponent  is  0  is'  equal  to  1. 

By  this  notation,  the  trace  of  a  letter  which  has  disappeared 
in  the  operation  of  division  may  be  preserved.  Thus,  the 
quotient  of  a2  b3  divided  by  a2  b2,  if  important  to  indicate  that 
a  originally  entered  into  the  term,  may  be  written  a0  b. 

95.  In  Division  we  may  distinguish  three  cases. 

CASE    I. 

96.  When  both  dividend  and  divisor  are  monomials. 
From  the  preceding  articles  is  derived  the  following 


DIVISION.  33 

RULE. 

Divide  the  coefficient  of  the  dividend  by  that  of  the  divisor  ; 
and  to  the  result  annex  the  letters  of  the  dividend,  each  with  an 
exponent  equal  to  its  exponent  in  the  dividend  diminished  by 
its  exponent  in  the  divisor;  omitting  all  letters  whose  expo- 
nents become  zero.  Make  the  quotient  +  when  the  dividend  and 
divisor  have  the  same  sign,  and  —  when  they  have  different 
signs. 

EXAMPLES. 

1.  Divide  9  a2  b  c  x  y  by  3  a  b  c. 

9a2bcxy-i-3abc  =  3axy,  Ans. 

2.  Divide  24  a4  m3  n2  by  —  8  a  m3  n. 

24  a4  m3  n2  -. —  8  a  m3  n  —  —  3  a3  n,  Ans. 

3.  Divide  —  35  xm  by  —  7  xn. 

—  35  xm  -^-  —  7xn  =  5  xm~n,  Ans. 

Divide  tbe  following : 

4.  12  a5  by  4  a.  8.    -  65  a3  b3  c3  by  -  5  a  b2  c3. 

5.  6  a2  c  by  6  a  c.  9.    72  m°  n  by  —  12  m2. 

6.  14  m3  n4  by  -  7  m  n3.  10.    -  144  c5  <P  e6  by  36  c2  d3  e. 

7.  -18x2y5zhy9x2z.  11.    -  91  x4  y3  z2  by  -  13  x3  y1. 

CASE    II. 

97.  When  the  dividend  is  a  polynomial  and  the  divisor  is 
a  monomial. 

Tbe  operation  being  just  tbe  reverse  of  that  of  Art.  85,  we 
have  the  following 

RULE. 

Divide  each  term  of  the  dividend  by  the  divisor,  remembering 
that  like  signs  produce  +,  and  tinlike  signs  produce  — . 


34  ALGEBRA. 

EXAMPLES. 

1.  Divide  9  a3  ft  +  6  a4  e  — 12  a  ft  by  3  a. 

3«)9a3H6  a4 c  — 12  a  5 
3a2i  +  2«3c-4J,    ^4»s. 

Divide  the  following : 

2.  8  a3  6  c  +  16  a5  6  c  —  4  a2  c2  by  4  a2  c. 

3.  9  a5  ft  c  -  3  a2  ft  +  18  a3  ft  c  by  3  a  ft. 

4.  20  a4  ft  c  +  15  a  6  d3  -  10  a2  ft  by  -  5  a  ft. 

5.  3  a3  (a  -  ft)  +  9  a  (a  +  ft)  by  3  a. 

6.  15  (x  +  y)2  —  5  a  (x  +  y)  +  10  ft  (x  +  y)  by  —  5  (x  +  y). 

7.  4  x7  -  8  a-6  - 14  t>  +  2  a4  -  6  x3  by  2  x\ 

8.  9  a4  +  27  x3- 21  a2  by -3  a;2. 

9.      _  a6  £6  C4  _  a4  &5  c3  +  3  a8  £4  ^2  fcy  _  ft8  p  ^ 

10.   —  12  aP  ft?  -  30  a12  ft3  +  108  a"  ftn  by  -  6  am  ft'". 

CASE     III. 

98.      When  the  divisor  is  a  polynomial. 

1.  Let  it  be  required  to  divide  12  +  10  x3  —  11  x  —  21  x2  by 
2z2-4-3x. 

We  are  then  to  find  a  quantity  which  when  multiplied  by 
2  x2  -  4  -  3  x  will  produce  12  +  10  x3  - 11  x  -  21  x2. 

Now,  in  the  product  of  two  polynomials,  the  term  containing 
the  highest  power  of  any  letter  in  the  multiplicand,  multiplied 
by  the  term  containing  the  highest  power  of  the  same  letter 
in  the  multiplier,  produces  the  term  containing  the  highest 
power  of  that  letter  in  the  product.  Hence,  if  the  term  con- 
taining the  highest  power  of  x  in  the  dividend,  10  x3,  be  di- 
vided by  the  term  containing  the  highest  power  of  x  in  the 
divisor,  2  x2,  the  result,  5  x,  will  he  the  term  containing  the 
highest  power  of  x  in  the  quotient. 


DIVISION.  35 

Multiplying  the  divisor  by  5  x,  the  term  of  the  quotient 
already  found,  and  subtracting  the  result,  10  x3  —  20  x  —  15  x2, 
from  the  dividend,'  the  remainder,  12  +  9  x  —  6  x'2,  may  be  re- 
garded as  the  product  of  the  divisor  by  the  rest  of  the  quotient. 

Therefore,  to  find  the  rest  of  the  quotient,  we  proceed  as  be- 
fore, regarding  12  +  9  x  —  6  x2  as  a  new  dividend,  and  divid- 
ing the  term  containing  the  highest  power  of  x,  —  6  x'2,  by 
the  term  containing  the  highest  power  of  x  in  the  divisor,  2  x2, 
giving  as  a  result  —  3,  which  is  the  term  containing  the  high- 
est power  of  x  in  the  rest  of  the  quotient. 

Multiplying  the  divisor  by  —  3,  the  term  of  the  quotient 
just  found,  and  subtracting  the  result,  —  6  x2  +  12  +  9  x,  from 
the  second  dividend,  there  is  no  remainder.  Hence,  5  x  —  3 
is  the  quotient  required. 

99.  It  will  be  observed  that  in  getting  the  terms  of  the 
quotient,  we  search  for  the  terms  containing  the  highest  power 
of  some  letter  in  the  dividend  and  divisor.  These  may  be 
obtained  most  conveniently  by  arranging  both  dividend  and 
divisor  in  order  of  powers  commencing  with  the  highest 
(Art.  41)  ;  this,  too,  facilitates  the  subsequent  subtraction. 
We  also  should  arrange  each  remainder  or  new  dividend  in 
the  same  order. 

It  is  customary  to  arrange  the  work  as  follows : 


10  x3  -  21  x2  -  11  x  +  12 
10  xs  -  15  x2  -  20  x 


2  x2  —  3  x  —  4,  Divisor. 
5  x  —  3,  Quotient. 


—  6x2  +    9  a: +  12 

-  6x2+    9  a; +  12 


100.  We  might  have  obtained  the  quotient  by  dividing  the 
term  containing  the  lowest  power  of  x  in  the  dividend,  12,  by 
the  term  containing  the  lowest  power  of  x  in  the  divisor,  —  4, 
which  would  have  given  as  a  result  — 3,  the  term  containing 
the  lowest  power  of  x  in  the  quotient.  In  solving  the  problem 
in  this  way,  we  should  first  arrange  both  dividend  and  divisor 
in  order  of  powers  commencing  with  the  lowest,  and  should 


o 


G  ALGEBRA. 


afterwards  bring  clown  each  remainder  in  the  same  order;  re- 
membering that  a  term  which  does  not  contain  x  at  all  con- 
tains a  lower  power  of  x  than  any  term  which  contains  x. 

101.     From  the  preceding  principles  we  derive  the  follow- 


ing 


RULE. 

Arrange  both  dividend  and  divisor  in  the  same  order  of  pow- 
ers of  some  common  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor,  and  write  the  result  as  the  first  term  of  the  quotient. 

Multiply  the  whole  divisor  by  this  term,  and  subtract  the 
product  from  the  dividend,  arranging  the  result  in  the  same 
order  of  powers  as  the  divisor  and  dividend. 

Regard  the  remainder  as  a  new  dividend,  and  divide  its  first 
term  by  the  first  term  of  the  divisor,  giving  the  next  term  of  the 
quotient. 

Multiply  the  whole  divisor  by  this  term,  and  subtract  the 
product  from  the  last  remainder. 

Continue  in  the  same  manner  until  the  remainder  becomes 
zero,  or  until  the  first  term  of  the  remainder  will  not  contain 
the  first  term  of  the  divisor. 

When  a  remainder  is  found  whose  first  term  will  not  con- 
tain the  first  term  of  the  divisor,  the  remainder  may  be  written 
with  the  divisor  under  it  in  the  form  of  a  fraction,  and  added 
to  the  quotient. 

2.    Divide  a8  -  3  a2  b  +.  12  bz  +  5  a  b2  by  b  +  a. 
Arranging  the  dividend  and  divisor  in  order  of  powers, 

a  +  b)  a3  -  3  a2  b  +  5  a  b2  +  12  bz  {a2  -  4  a  b  +  9  b2 
a3  +  a2  b 

-  4  a2  b  ~ 

—  4  a2  b  —  4  a  b2 

9ab2 

9  a  b2  +9bs 


3  b3,  Remainder. 

3  b9 


Ans,    a2-4ab  +  9b2  + 


a  +  b' 


DIVISION.  37 

EXAMPLES. 

3.  Divide  2  a2  x2  —  5  a  x  +  2  by  2  a  x  —  1. 

4.  Divide  3  63  +  3  a  b2  -  4  a2  b  -  4  «3  by  a  +  b. 

5.  Divide  8  a?  -  4  a2  i  -  6  a  b2  +  3  &3  by  2  a  -  J. 

6.  Divide  21  a5  -  21  b5  by  la  — lb. 

7.  Divide  a3  +  2  a;3  by  a  +  #. 

8.  Divide  x*  +  y4  by  cc  +  y. 

9.  Divide  23  x"- 48  +  6  cc4 -  2  a;  -  31  x*  by  6  +  3a;2  —  5  x. 

10.  Divide  15  x4-  32  x3  +  50  x2  -  32  a;  +  15  by  3  x2-  4  x  +  5. 

11.  Divide  2  a;4  -  11  aj  -  4  ar  -  12  -  3  x3  by  4  +  2  ar  +  jb. 

12.  Divide  a;5  —  v/5  by  x  —  y. 

13.  Divide  35  - 17  x  +  16  x2  -  25  a;3  +  6  x4  by  2  x  —  1. 

14.  Divide  3  x2  +  4  x  +  6  a;0  -  11  x3  -  4  by  3  x2  -  4. 

15.  Divide  a2  —  &'2  +  2  &  c  —  c2  by  a  +  b  —  c. 

16.  Divide  a;4  —  9  a;2  —  6  x  y  —  y2  by  x2  +  3  x  +  y. 

17.  Divide  xn  +  1  +  xn  y  +  x  yn  +  yn  + 1  by  xn  +  yn. 

18.  Divide  crn  —  b2m  +  2  bm  cr  —  c2r  by  an  +  bm  —  cr. 

19.  Divide  1  +  a  hy  1  —  a. 

In  examples  of  this  kind  the  division  does  not  terminate, 
there  being  a  remainder  however  far  the  operation  may  be 
carried. 

20.  Divide  a  by  1  +  x. 

21.  Divide  a8  +  a6  b2  +  a4  b4  +  a2b*  +  ba 

by  a4  +a3b  +  a2  b2  +  ab3  +  b4. 

22.  Divide  3  a3  +  2  -  4  a5  +  7  a  +  2  a6  -  5  a4  +  10  a2 

by  a3  — 1  — a2  — 2  a. 

23.  Divide  15  x2  -  x4  -  20  -  2  a;5  +  6  x  +  2  x3 

by  5  -  3  a;2  -  4  x  +  2  x3. 


38  ALGEBRA. 

24.  Divide  2  x5  +  4  x2  — 14  +  7  x  —  7  x3  +  x«  —  xi 

by  2  x2  -  7  +  z3. 

25.  Divide  12  «5  -  14  a4  b  +  10  a8  i2  -  a2  63  -  8  a  i4  +  4  i5 

by  6  a8  -  4  a2  6  -  3  a  b2  +  2  63. 

102.  In  Art.  88  we  showed  that  0  X  a  =  0.  Since  the 
product  of  the  divisor  and  quotient  equals  the  dividend,  we 
may  regard  0  as  the  quotient,  a  as  the  divisor,  and  0  as  the 
dividend.     Therefore, 

°-  =  0. 

That  is,  a 

If  zero  be  divided  by  any  quantity  the  quotient  is  equal  to 
zero. 


VII.  — FORMULAE. 

103.  A  Formula  is  an  algebraic  expression  of  a  general  rule. 
The  following  formulae  will  be  found  very  useful  in  abridg- 
ing algebraic  operations. 

104.  By  Art.  17,  (a  +  bf  =5  (a  +  b)  (a  +  b)  ;  whence,  by 
actual  multiplication,  we  have 

That  is,  (a  +  V*  =  cc2  +  2ab  +  b2.  (1) 

The  square  of  the  sum  of  two  quantities  is  equal  to  the 
square  of  the  first,  plus  twice  the  product  of  the  first  by  the 
second,  plus  the  square  of  the  second. 

105.  We  may  also  show,  by  multiplication,  that 

(a  -  b)2  =  a2-2ab  +  b\  (2) 

That  is, 

The  square  of  the  difference  of  two  quantities  is  equal  to 

the  square  of  the  first,  minus  twice  the  product  of  the  first  by 

the  second,  plus  the  square  of  the  second. 

106.  Again,  by  multiplication,  we  have 

(a  +  b)  (a-b)  =  a2-  b2.  (3) 


FORMULAE.  3y 

That  is, 

The  product  of  the  sum  and  difference  of  two  quantities  is 
equal  to  the  difference  of  their  squares. 

EXAMPLES. 

1     107.     1.    Square  3  a  +  2  b. 

The  square  of  the  first  term  is  9  a2,  twice  the  product  of  the 
terms  is  12  a  b,  and  the  square  of  the  last  term  is  4  b2.  Hence, 
by  formula  (1), 

(3  a  +  2  b)2  =  9  a2  +  12  a  b  +  4  b2,  Am. 
Square  the  following : 

2.  2m +  3 w.  4.  3 x  +  11.  6.  2ab  +  &ac. 

3.  x2  +  4.  5.  4a  +  5 b.  7.  7 x3  +3x. 
8.    Square  4  x  —  5. 

The  square  of  the  first  term  is  16  x2,  twice  the  product  of 
the  terms  is  40  a-,  and  the  square  of  the  last  term  is  25. 
Hence,  by  formula  (2), 

(4  x  -  5)2  =  16  x2  -  40  x  +  25,  Am. 

Square  the  following : 

9.  3a2-bs.  11.  l-2£  13.  3-a\ 

10.  4  a  b  -  x.  12.  x*  -  y\  14.  2  a;3  -  9  x2. 

15.   Multiply  6  a  +  b  by  6  «  —  6. 

The  square,  of  the  first  term  is  36  a2,  and  of  the  last  term  b2. 
Hence,  by  formula  (3), 

(6a  +  b)  (6  a  -  b)  =  36  a2  -  b2,  Am. 
Expand  the  following : 

16.  (x  +  3)  (x  -  3).  19.   (am  +  a")  (am  -  an). 

17.  (2  x  +  1)  (2  x  -  1).  20.   O3  +  5  a-)  (a;3  -  5  a-). 

18.  (5«  +  7J)  (5a-76).      21.  (4ar  +  3)  (4a;2-3). 
22.    Multiply  a  +  b  +  chja  +  b  —  c. 

(a  +  b  +  c)  (a  +  b  -  c)  =  [(»  +  b)  +  c]  [O  +  6)  -  c] 
=  (Art.  106),  (a  +  b)2  -  c2=  a2  +  2  «  b  +  b2  -  c2,  Am. 


40  ALGEBRA. 

Expand  the  following : 

23.  [1  +  («-&)]  [1-  («-  &)]•     25-  (a-5  +  c)(a  — &-c). 

24.  (a  +  &  +  c)  (a  —  &  —  c).  26.  (c  +  a  -  &)  (c  —  a  +  b). 

27.  [(a  +  b)  +  (c-d)2  [(a  +  ft)  -  («-«*)]. 

28.  (a  —  b  +  c  —  d)(a  —  b  —  c  +  d). 

29.  (a  +  b  +  c  +  d)  (a  +  b  —  c  —  d). 


VIII.  — FACTORING. 

108.  The  Factors  of  a  quantity  are  such  quantities  as  will 
divide  it  without  a  remainder. 

109.  Factoring  is  the  process  of  resolving  a  quantity  into 
its  factors. 

110.  A  Prime  Quantity  is  one  that  cannot  be  divided, 
without  a  remainder,  by  any  integral  quantity,  except  itself 
or  unity. 

Thus,  a,  b,  and  a  +  c  are  prime  quantities. 
Quantities  are  said  to  he  prime  to  each  other  when  they  have 
no  common  integral  divisor  except  unity. 

111.  One  quantity  is  said  to  be  divisible  by  another  when 
the  latter  will  divide  the  former  without  a  remainder. 

Thus,  a  b  and  a  b  +  a2  b'1  are  both  divisible  by  a,  b,  and  a  b. 

112.  If  a  quantity  can  be  resolved  into  two  equal  factors, 
it  is  said  to  be  a, perfect  square  ;  and  one  of  the  equal  factors 
is  called  the  square  root  of  the  quantity. 

If  a  quantity  can  be  resolved  into  three  equal  factors,  it  is 
said  to  be  a  perfect  cube  ;  and  one  of  the  equal  factors  is  called 
the  cube  root  of  the  quantity. 

Thus,  since  1  a-  equals  2  a  X  2  a,  4  a2  is  a  perfect  square 
and  2  a  is  its  square  root ;  and  since  27  Xs  equals  3,xx3a,x3/, 
27  Xs  is  a  perfect  cube,  and  3x  is  its  <nbe  root. 


FACTORING.  41 

Note.  4  a2  also  equals  -  2  a  x  -  2  a,  so  that  the  square  root  of  4  <>2 
may  be  either  2  a  or  -  2  «.  In  the  examples  in  this  chapter  we  shall  only 
consider  the  positive  square  root. 

To  find  the  square  root  of  an  algebraic  quantity,  extract  the  square  root 
of  the  numerical  coefficient,  and  divide  the  exponent  of  each  letter  by  2. 
Thus,  the  square  root  of  9  a6  b2  is  3  a?  b. 

To  find  the  cube  root,  extract  the  cube  root  of  the  numerical  coefficient, 
and  divide  the  exponent  of  each  letter  by  3.  Thus,  the  cube  root  of  8  a3  b® 
is  2  a  b2. 

113.  The  factoring  of  monomials  may  be  performed  by- 
inspection  ;  thus, 

12  a3  b2  c  =  2.2.8.  a  a  abbe. 
But  in  the  decomposition  of  polynomials  we  are  governed  by 
rules  which  may  be  derived  from  the  laws  of  their  formation. 
A  polynomial  is  not  always  factorable ;  but  in  numerous  cases 
we  can  always  factor ;  and  these  cases,  together  with  the  rules 
for  their  solution,  will  be  found  in  the  succeeding  articles. 

CASE    I. 

114.  Wlien  the  terms  of  a  polynomial  have  a  common  mo- 
nomial factor,  it  may  be  written  as  one  of  the  factors  of  the 
polynomial,  ivith  the  quotient  obtained  by  dividing  the  given 
polynomial  by  this  factor,  as  the  other. 

1.    Factor  the  expression  3  x3  y2  —  12  x  y4  —  9  x2  y3. 
We  observe  that  each  term  contains  the  factor  3  x  y2. 
Dividing  the   given   polynomial   by  3  x  y2,  we  obtain  as  a 
quotient  x2  —  4  y2  —  3  x  y.     Hence, 

3 x3 y2 -  12  x  y4  -  9  x2 yz  =  3  .« y2  (x2-±y2-3x  y),  Arts. 

EXAMPLES. 

Factor  the  following  expressions  : 

2.  as  +  a.  $  5.  60m4  n2  — 12  n\ 

3.  16  x4  - 12  x.  6.  27  c4  d2  +  9  c3  d. 

4.  a& -2  a4 +  3  a3 -a2.  7.  36  Xs  y  —  60  x2  y4  —  84  x4  y2. 

8.  a5b-3a6b4-2a3b4c  +  6a'1b5x. 

9.  84  x2  y3  -  140  x3  y4  +  56  x4  if. 


42  ALGEBRA. 

10.  72  o4  b3  c3  +  126  a3  c2  d  +  162  a2  c. 

1 1.  128  c4  d5  +  320  c2  d7  -  448  c8  <Z4. 

CASE    II. 

115.      TFAerc  a  polynomial  consists  of  four  terms,  the  first 
two  and  last  two  of  which  have  a  common  binomial  factor,  it 

may  he  written  as  one  of  the  factors  of  the  polynomial,  wit  li- 
the quotient  obtained  by  dividing  the  given  polynomial  by  this 
factor,  as  the  other. 

1.  Factor  the  expression  a  m  —  b  m  +  a •  n  —  b  n. 
Factoring  the  first  two  and  last  two  terms  by  the  method 

of  Case  I,  we  obtain  m  (a  —  b)  +  n  (a-  —  b).  We  observe  that 
the  first  two  and  last  two  terms  have  the  common  binomial 
factor  a  —  b.  Dividing  the  expression  by  this,  we  obtain  as  a 
quotient  m  +  n. 

Hence,  am  —  bm  +  an  —  bn=(a  —  b)  (m  +  n),  Ans. 

2.  Factor  the  expression  a  m  —  bm  —  an  +  b  n. 

am  —  b  7)i  —  a  n  +  b  n  =■  a  m  —  b  m  —  (a  n  —  b  11)  =  ni  (a  —  b) 
—  n(a  —  b)  —  (a  —  b)  (m  —  n),  Ans. 

Note.  If  the  third  term  of  the  four  is  negative,  as  in  Ex.  2,  it  is 
convenient  to  enclose  the  last  two  terms  in  a  parenthesis  with  a  -  sign 
prefixed,  before  factoring. 

EXAMPLES. 

Factor  the  following  expressions  : 

3.  a  b  +  b  x  +  a  y  +  x  y.  -7.  mx2  —  my2  —  ?ix2+  n  y1. 

4.  a  c  —  cm  +  a  d  —  dm.  8.  x3  +  x2  +  x  +  1. 

5.  x2  +  2x  —  xy  —  2y.  9.  6  x3  +  4  x2  —  9  x  -  6. 

6.  a3  —  a2b  +  a  b2  -  b3.  10.  8  c  x  -  12  c  y  +  2  d  x  -  3  d  y. 

11.  6  n  -  21  m2  n-8m  +  28  m3. 

12.  a2bc  —  ac2d+ab2d  —  bc  d2. 

13.  m2n2  x2  —  ns  x  y  —  m3  x  y  +  m  n  y'2. 

14.  12  a  b  m  n  —  21  a  b  x  y  +  20  <■  d  m  n  —  35  c  d  x  //. 


FACTORING.  4:3 

CASE    III. 

116.  When  the  first  and  last  terms  of  a  trinomial  are 
perfect  squares  and  positive,  and  the  second  term  is  twice  the 
product  of  their  square  roots. 

Comparing  with  Formulae  1  and  2,  Arts.  104  and  105,  we 
observe  that  such  expressions  are  produced  by  the  product  of 
two  equal  binomial  factors.  Reversing  the  rules  of  Arts.  104 
and  105,  we  have  the  following  rule  for  obtaining  one  of  the 
equal  factors : 

Extract  the  square  roots  of  the  first  and  last  terms,  and 
connect  the  results  by  the  sign  of  the  second  term. 

1.  Factor  a2  +  2  a  b  +  b2. 

The  square  root  of  the  first  term  is  a ;  of  the  last  term,  b ; 
the  sign  of  the  second  term  is  + .  Hence,  one  of  the  equal 
factors  is  a  +  b. 

Therefore,  a2  +  2  a  b  +  b2  =  (a  +  b)  (a  +  b)  or  (a  +  b)2,  Ans. 

2.  Factor  9  or  —  12  a,  b  +  A  hi 

The  square  root  of  the  first  term  is  3  a  ;  of  the  last  term,  2  b ; 
the  sign  of  the  second  term  is  — .  Hence,  one  of  the  equal 
factors'  is  3  a  —  2  b.     Therefore, 

9  a2- 12  a  b  +  4  b2  =  (3  a  -2  b)  (3  a  -  2  b)  or  (3  a  -  2  bf,  Ans. 

Note.     According  to  Art.    58,    the   given    expression   may  be   written 
4  b2  —  12  a  b  +  9  a2.     Applying  the  rule  to  this  expression,  we  have 

4  b2-  12  a  b  +  9  a2  =  (2  b-  3  a)  (2  b  -  3  a)  or  (2  b  -  3  a)2. 

We  should  obtain  this  second  form  of  the  result  in  another  way  by  apply- 
ing the  principles  of  Art.  89  to  the  first  factors  obtained. 

EXAMPLES. 

Factor  the  following  expressions  : 

3.  x2  - 14  x  +  49.  6.  a2  -  28  a  +  196.. 

4.  m2  +  36  m  +  324.  7.  n6  -  26  n3  +  169. 

5.  y2  +  20  y  +  100.  8.  x2  y2  +  32  x  y  +  256. 


44  ALGEBRA. 

9.  25  x2  +  70  x  y  z  +  49  y2  z2.  11.  16  m2  -8am+«2. 

10.  36  m?  -  36  w  «  +  9  n2.  12.  4  a2  +  44  a  b  +  121  b2. 

13.  a2  64  +  12  a  b2  c  +  36  c2. 

14.  9  «4  +  60  a2  bc2d  +  100  b2  ci  d2. 

15.  4  xA  -  60  m  re  x2  +  225  m2  t»s 

16.  64  x6  -  160  x5  +  100  cc4. 

CASE    IV. 

117.  When  an  expression  is  the  difference  between  two 
perfect  squares. 

Comparing  with  Formula  3,  Art.  106,  we  observe  that  such 
expressions  are  the  product  of  the  sum  and  difference  of  two 
quantities.  Reversing  the  rule  of  Art.  106,  we  have  the  fol- 
lowing rule  for  obtaining  the  factors  : 

Extract  the  square  roots  of  the  first  and  last  terms  ;  add 
the  results  for  one  factor,  and  subtract  the  second  result  from 
the  first  for  the  other.  • 

1.  Factor  36  x2  —  49  y2. 

The  square  root  of  the  first  term  is  6  a;;  of  the  last,  7  y. 
The  sum  of  these  is  6  x  +  7  y,  and  the  second  subtracted  from 
the  first  is  6  x  —  7  y.     Hence, 

36  x2  —  49  y2  =  (6  x  +  7  y)  (6  x  —  7  y),  Ans. 

2.  Factor  {a  -  b)2  -  (c-df. 

The  square  root  of  the  first  term  is  a  —  b  ;  of  the  last,  c  —  d. 
The  sum  of  these  is  a  —  b  +  c  —  d,  and  the  second  subtracted 
from  the  first  is  a  —  b  —  c  +  d.     Hence, 

(a  —  b)2  —  (c  —  d)2  =  (a  —  b  +  c  —  d)  (a  —  b  —  c  +  cl),  Ans. 

EXAMPLES. 

Factor  the  following  expressions  : 

3.  x2-l.  5.  a4 -/A  7.  4 a*  —  225 ma w9. 

4.  4x2-9t/2.  6.  9  a2 -4.  8.  1  -  196  x2  if  z\ 


FACTORING.  45 

9.   (a  +  b)2  -  (c  +  d)2.  11.  m2  -  (x  -  y)2. 

10.  (a-c)*—b*.  12.  (aj  — m)a— (y  — »)*. 

Many  polynomials,  consisting  of  four  or  six  terms,  may  be 
expressed  as  the  difference  between  two  perfect  squares,  and 
hence  may  be  factored  by  the  rule  of  Case  IV. 

13.  Factor  2  m  n  +  m2  —  1  +  n2. 

Arrange  the  expression  as  follows,  m2  +  2mn  +  r?  —  1. 
Applying  the  method  of  .Case  III  to  the  first  three  terms,  we 
may  write  the  expression  (m  +  n)'2 —  1.  The  square  root  of 
the  first  term  is  m  +  n ;  of  the  last,  1.  The  sum  of  these  is 
m  +  n  +  1,  and  the  second  subtracted  from  the  first  is 
m  +  n  —  1.     Hence, 

2  m  n  +  m2  —  1  +  n2  =  (m  +  n  +  1)  (m  +  n  —  1),  Ans. 

14.  Factor  2  x  y  +  1  —  x2  —  y2. 

2  x  y  +  1  —  x2  —  y2  =  l  —  x2  -\-2x  y  —  y2 
=  1-  {x2 -2  x  y  +  y2)  =  l  -  (x-y)2,  by  Case  III, 
=  [1  +  (x— y)][l  —  (as— y)]  =  (l  +  #-y)  (1—  x  +  y),Ans. 

15.  Factor  2xy+b2-x2-2ab-  y2  +  a2. 

2xy+b2  —  x2  —  2ab  —  y2+  a2 
=  a2  —  2ab  +  b'2  —  x2  +  2  x  y  —  y2 
=  a2-2  a  b  +  b2  -  (xs  -  2  x  y  +  y2) 
=  (a  -  b)2  —  (x—  y)2,  by  Case  III, 
=  [(«-&)+ (as-y)][(a-fl)-(aj-y)] 
=  (a  —  b+x  —  y)  (a—b  —  x  +  y),  Ans. 

Factor  the  following  expressions  : 

16.  x2  +  2xy  +  y2-4:.  19.  9-x4-4?/2  +  4f2y. 

17.  aa  _  j2  +  2  h  c  _  c2.  20.  4  a2  +  62  -  9  d?  -4ab. 

18.  9  c2  - 1  +  d?  +  6  c  d.  21.  4  b  - 1  -  4  b2  +  4  m4. 

22.  a2  —  2  a  m  +  m2  —  b2  —  2  b  n  —  n2. 

23.  2  a  m  —  62  +  m2  +  2bn+  a2  -  n2. 

24.  x2  -  y2  +  c2  -  d2  -  2  c  x  +  2  tf  y. 


46  ALGEBRA. 


CASE   V. 


118.  Wlien  an  expression  is  a  trinomial,  of  the  form 
x-  +  a  x  +  b  ;  where  the  coefficient  of  x2  is  unity,  and  a  and 
b  represent  any  whole  numbers,  either  positive  or  negative. 

To  derive  a  rule  for  this  case  we  will  consider  four  examples 
in  Multiplication : 

I.  ii. 

x  +  5  x  —  o 

x  +  3  x  —  3 


x1  +  5  x  x2—ox 

+  3.x- +  15  -3cc  +  15 

a;2  +  Sx  +  15  sca-8aT+15 

III.  IV. 

x  +  5  x  —  5 

x  —  3  x  +  3 


a?2  +  5  x  #2  —  5  .<■ 

-3x-15  +3x-ll 


«2+2^-15  x2-2ic-15 

We  observe  in  these  results, 

1.  The  coefficient  of  x  is  the  algebraic  sum  of  the  numbers 
in  the  factors. 

2.  The  last  term  is  the  product  of  the  numbers. 

Hence,  in  reversing  the  process,  we  have  the  following  rule 
for  obtaining  the  numbers  : 


RULE. 

Find  two  numbers  ivhose  algebraic  sum  is  the  coefficient  of 
x,  and  whose  product  is  the  last  term. 

Note.     We  may  shorten  the  work  by  considering  the  following  points  : 

1.  When  the  last  term  of  the  product  is  f,  as  in  Examples  I  ami  II, 
the  sum  of  the  numbers  is  the  coefficient  of  .<• ;  both  numbers  being  + 
when  the  second  term  is  +,  and  -  when  the  second  term  is  -. 

2.  When  the  last  term  is  -,  as  in  Examples  III  and   I  V,  the  difference 


FACTORING.  47 

of  the  numbers  (disregarding  signs)  is  the  coefficient  of  x ;  the  larger 
number  being  +  and  the  smaller  -  when  the  second  term  is  +,  and  the 
larger  number  -  and  the  smaller  +  when  the  second  term  is  -  . 

We  may  embody  these  observations  in  the  form  of  a  rule  which  may  be 
found  more  convenient  than  the  preceding  rule  in  the  solution  of  examples. 

I.  If  the  last  term  is  +,  find  tivo  members  whose  sum  is  the  coefficient  of 
x,  and  whose  product  is  the  last  term;  and  give  to  both  numbers  the  sign  of 
the  second  term. 

II.  If  the  last  term  is  -  ,  find  tivo  numbers  whose  difference  is  the  coeffi- 
cient of  x,  and  whose  product  is  the  last  term;  and  give  to  the  larger  num- 
ber the  sign  of  the  second  term,  and  to  the  smaller  number  tlie  opposite  sign. 

1.  Factor  x2  +  14  x  +  45. 

Here  we  are  to  find  two  numbers  whose  -  "       - 

(.product  —  45  J 

The  numbers  are  9  and  5 ;  and,  the  second  term  being  + ,  both 
numbers  are  +.     Hence, 

x2  +  14  x  +  45  =  (x  +  9)  (x  +  5),  Ans. 

2.  Factor  x2  —  6  x  +  5. 

Here  we  are  to  find  two  numbers  whose  ■!  \ 

(product  =  5,' 

The  numbers  are  5  and  1 ;  and,  as  the  second  term  is  — ,  both 
numbers  are  — .     Hence, 

x2  —  6  x  +  5  =  (x  —  5)  (x  —  1),  Ans. 

3.  Factor  x2  +  5  x  —  14. 

Here  we  are  to  find  two  numbers  whose  ■<  " 

I  product     =  14 ) 

The  numbers  are  7  and  2;  and  as  the  second  term  is  +,  the 
larger  number  is  + ,  and  the  smaller  — .     Hence, 

x2  +  5  x  —  14  =  (x  +  7)  (x  —  2),  Ans. 

4.  Factor  x2  —  7  x  —  30. 

Here  we  are  to  find  two  numbers  whose  ]  ~       \ 


48  ALGEBKA. 

The  numbers  are  10  and  3;  and  as  the  second  term  is  — ,  the 
larger  number  is  — ,  and  the  smaller  + .     Hence, 

x2  -  7  x  -  30  =  (x  - 10)  (x  +  3),  Arts. 

Note.  In  case  the  numbers  cannot  be  readily  determined  by  inspection, 
the  following  method  will  always  give  them  : 

Eequired  two  numbers  whose  difference  is  8  and  product  48.     Taking  in 

order,  beginning  with  the  lowest,  all  possible  pairs  of  integral  factors  of  48, 

we  have 

1x48, 

2x24, 

3x16, 

4x12. 

And,  as  4  and  1 2  differ  by  8,  they  are  the  numbers  required. 

Evidently  this  method  will  give  the  required  numbers  eventually,  how- 
ever large  they  may  be,  provided  they  exist. 

EXAMPLES. 

Factor  the  following  expressions  : 

5.  a2 +5  a- +  6.  12.  m2+9m  +  8. 

6.  a2-3a  +  2.  13.  m'2  +  2m-80. 

7.  2/2  +  2?/- 8.  14.  c2- 18  c +  32. 

8.  m2-5m-24.  15.  x2  +  x-42. 

9.  *2-ll;c  +  l8.  16.  x2  +  23x  +  102. 

10.  n*-n-c)0.  17.   if-9y-90. 

11.  *2+13a;  +  36.  18.   a2+13a-48. 

19.   cc2-9z-70. 

20.   Factor  15  —  2x  —  x2. 

15  _  2  x  -  x2  =  -  (x2  +  2  x  - 15) 

By  the  rule  of  Case  V,  x2  +  2  x  -  15  =  (.r  +  5)  (x  -  3). 

Hence, 

15  -  2  x  -  x2  =  -  (x  +  5)  (x  -  3)  =  (x  +  5)  (3  -  x),  Ans. 

Note.  If  the  x*  term  is  -,  enclose  the  whole  expression  in  a  paren- 
thesis with  a  -  sign  prefixed.  Factor  the  quantity  within  the  parenthesis, 
and  change  the  signs  of  all  the  terms  of  one  factor. 


FACTORING.  49 

Factor  the  following  expressions  : 
21.  20-x-x2.  22.  6  +  x-x2.  23.   84-5z-a;2. 

The  method  of  Case  V  may  he  extended  to  the  factoring  of 
more  complicated  trinomials. 

24.    Factor  m2 ri2  —  3  m  nx  +  2  x2. 

r.    i  it         (sum         =3) 

Here  we  are  to  find  two  numbers  whose  <  t  —  9  j 

The  numbers  are  2  and  1;  and  as  the  second  term  is—, 
hoth  numbers  are  — .     Hence, 

m2 ri2  —  3  m nx  +  2x2  =  (m n  —  2 x)  (m n  —  x),  Ans. 

Factor  the  following  expressions  : 

25.  a-4  -  29  x2  +  120.  30.   ?«4  +  5  m2  n2  -  66  n\ 

26.  c6  +  12c3+ll.  31.    (a-b)2-S(a-b)-4. 

27.  x2yG  +  2xf-120.  32.    (x  +  y)*-  7  (x  +  y)  +  10. 

28.  (r'^_7«i2-144.  33.   x2 -  2  x  y2  z  -  48  y*  z\ 

29.  x2  +  25  w  x  +  100  ?i2.  34.    (m  +  nf  +  (m  +  n)  -  2. 

CASE   VI. 

119.     When  an  expression  is  the  sum  or  difference  of  two 
perfect  cubes. 

By  actual  division,  we  may  show  that 

a3  +  b3  a3  —  b3 

— —  =  a2  —  ab  +  b'2,  and —  =  a2  +  a  b  +  b2. 

a  +  b  a  —  b 

Whence, 

(a3  +  b3)  =  (a  +  b)  (a2  -ab  +  b2),  and 

(a3  -  b3)  =  (a  -  b)  (a2  +  ab  +  b2). 

These  results  may  he  enunciated  as  follows  : 

To  factor  the  sum  of  two  perfect  cubes,  write  for  the  first 
factor  the  sum  of  the  cube  7'oots  of  the  quantities;  and  for  the 


50  ALGEBEA. 

second,  the  square  of  the  first  term  of  the  first  factor,  minus 
the  product  of  the  two  terms,  phis  the  square  of  the  last  term. 
To  factor  the  difference  of  two  perfect  cubes,  write  for  the 
first  factor  the  difference  of  the  cube  roots  of  the  quantities  ; 
and  for  the  second,  the  square  of  the  first  term  of  the  first 
factor,  plus  the  product  of  the  two  terms,  plus  the  square  of 
the  last  term. 

1.  Factor  8  a3  +  1. 

The  cube  root  of  the  first  term  is  2  a ;  of  the  last  term,  1. 
Hence,  8  a3  +  1  =  (2  'a  +  1)  (4  a2  -  2  a  +  1),  Ans. 

2.  Factor  27  x6  -  64  y3. 

The  cube  root  of  the  first  term  is  3  x2 ;  of  the  last  term,  4  y. 
Hence, 

27  xG  -  64  y*  =  (3  x2  -  4  y)  (9  xi  +  12  x2  y  +  16  y2),  Ans. 

EXAMPLES. 

Factor  the  following  expressions  : 

6.  Sc6-d9.  9.  343 +  8  a3. 

7.  125  as-  216  m3.  10.  27  a;3 -125. 

8.  729  c3  dP  +  512.  11.  1000  -27  a3  b\ 

CASE   VII. 

120.      When  an  expression  is  the  sum  or  difference  of  two 
like  powers  ofttvo  quantities.  ■ 

The  following  principles  are  useful  to  remember : 

1.  an  —  bn  is  always  divisible  by  a  —  b,  if  n  is  an. integer. 

2.  an  —  bn  is  always  divisible  by  a  +  b,  if  n  is  an  even  integer. 

3.  an  +  bn  is  always  divisible  by  a  +  b,  if  n  is  an  odd  integer. 

We  may  prove  the  first  principle  as  follows : 
Commencing  the  division  of  an  —  bn  by  a  —  b,  wv  have 


3. 

iC3  —  y3. 

4. 

a3 +  8. 

5. 

m3  +  64  ?z,6. 

a"  —  bn 


FACTORING.  51 

a  —  b 


an    '  +  . . .  Quotient. 


- 


an    ]  b  —  bn       Remainder. 

an  —  bn  ,       an-}b  —  bn  ,        b  (an~l  —  bn~v) 

or,  -     — -  =an~i  H ; =  an~l  -\ '-, 

a  —  b  a  —  b  a  —  b 

It  is  evident  from  this  result  that,  if  a""1  —  &n_1  is  exactly 
divisible  by  a  —  b,  the  dividend  an  —  bn  will  be  exactly  divisi- 
ble by  a,  —  b.  That  is,  if  the  difference  of  two  like  powers  of 
two  quantities  is  divisible  by  the  difference  of  those  quantities, 
then  the  difference  of  the  next  higher  powers  of  the  same 
quantities  is  also  divisible  by  the  difference  of  the  quantities. 
But  as—  b3  is  divisible  by  a  —  b,  hence  a4—  b4  is  ;  and  since 
c^  —  b*  is  divisible  by  a  —  b,  a5  —  b5  is;  and  so  on  to  any 
power.     This  proves  the  first  principle. 

Similarly  the  second  and  third  principles  may  be  proved. 

By  continuing  the  division,  we  should  find, 

=  a"-1  +  an~2  b  +  an~3  b~  + +  abn~2  +  bn-1    (1) 

an~'2b  +  an-sb2— +  abn~2  —  fi""1'  (2) 


a 

-b 

10 

an 
a 

—  bn 

+  b 

= 

an- 

-l 

an 

L. 

+  bn 

— 

an~ 

-l 

a  +  b 


a 


2b  +  an-3b2- —abn-2+bn~1   (3) 


It  is  useful  to  remember  that  when  a  —  b  is  the  divisor,  all 
the  terms  of  the  quotient  are  +  ;  where  a  +  b  is  the  divisor, 
the  terms  of  the  quotient  are  alternately  +  and  — ,  the  last 
term  being  +  if  n  is  odd,  and  —  if  n  is  even., 

1.    Factor  a1  —  IP. 
Putting  n  —  7  in  (1),  we  have 


a1  -b 


a*  +  a5b  +  a4  b2  +  a3  b3  +  a2  b*  +  abs+  b6. 


a  —  b 
Hence, 

a7  -  V  =  (a  -  b)  (a6  +  a5  b  +  a4  b2  +  a3b3  +  a2  b4  +  a  b5  +  b6), 

Ans. 


52  ALGEBRA. 

2.  Factor  m5  +  x5. 

Putting  a  =  m,  b  =  x,  n==  5,  in  (3),  we  have 

1YI     -f-  X  .  „  2       2  3,4 

=  m  —  m°  x  +  m  x  —  mxA  +  x\ 

m  +  x 

Hence, 

m5  +  xs=  (m  +  x)  (ra4  —  m3  a;  +  ra2  a:2  —  m  x3  +  x4),  Ans. 

3.  Factor  x6  —  y6. 

Putting  a  =  x,  b  =  y,  n  =  6,  in  (1),  we  have 

x6 ifi 

—  =  xs  +  x*  y  +  xs  y2  +  x2  y3  +  x  y*  +  y5. 

Hence, 

x6  —  y6=  (x  —  y)  (x5  +  x4  y  +  xs  y2  +  x2  ys  +  x  y*  +  f),  Ans. 

Or,  putting  a  =x,  b  =  y,  ?i  =  6,  in  (2),  we  have 
x6 ?/6 

J_  —  ^5 ~,i  »,    I     ~,3  „fi         ™2  „.3    i_  ™  ,.4         „.5 


Hence, 


=  x°  —  a;*  ?/  +  x"  y  —  x"  y6  +  x  if  —  y" 


x6  —  y6=(x  +  y)  (xh  —  x4y  +  xs  y2  —  x2y3  +  xyi  —  ys),  Ans. 

EXAMPLES. 

Factor  the  following  expressions  : 

4.  x5  +  y5.         6.   to6-c6.  8.   ms -n*.  10.   a4 -16. 

5.  (*-d\  7.    a'  +  V.  9.    c7-l.  11.    a7 +128. 

121.  By  the  application  of  one  or  more  of  the  given  rules 
for  factoring,  a  quantity  may  sometimes  be  separated  into 
more  than  two  factors. 

1.    Factor  2  a  xs  y2  -  8  a  x  y\ 

By  Case  I,  2  a  x3  if  -  8  a  x  if  =  2  a  x  y2  (x2  -  4  if). 
Factoring  the  quantity  in  the  parenthesis  by  Case  IV, 
2  a  x3  if  —  8  a  x  if  =  2  a  x  y2  (x  +  2  y)  (x  —  2  y),  Ans. 

Note.  If  the  method  of  Case  I  is  to  be  used  in  connection  with  other 
cases,  it  should  be  applied  first. 


GREATEST   COMMON   DIVISOR.  53 

2.    Resolve  aG  —  b6  into  four  factors. 
By  Case  IV,  a6  -  bG  =  (a3  +  b3)  (a3  -  b3). 
By  Case  VI,  a3  +  b3  =  (a  +  b)  (a2  -ab  +  b2), 
and  a3  -  b3  =  (a  -  b)  (a2  +  ab  +  b2). 
Hence, 
a6  -bG=(a  +  b)  (a-  b)  (a2  -ab  +  b2)  (a2  +ab  +  b2),  Ans. 

EXAMPLES. 

Factor  the  following  expressions  : 

3.  3  a3  b  +  12  a2  b  +  12  a  b.  7.  3  «4  -  21  a3  +  30  a: 

4.  45  x3  if  — 120  x2  f  +  80  x  if.      8.  2  c3  m  +  8  c2ra-42  c  m. 

5.  18  x3  y  —  2  x  f.  9.  m2a;y-4wa;y— 12 ay. 

6.  x3  +  Sx2+7x.  10.  32  «4  b  +  4  a  64. 

11.  Resolve  n9  —  1  into  three  factors. 

12.  Resolve  xi  —  if  into  three  factors. 

13.  Resolve  x8  —  m8  into  four  factors. 

14.  Resolve  m6  —  ?i6  into  four  factors. 

15.  Resolve  a9  +  c9  into  three  factors. 

16.  Resolve  64  «6  —  1  into  four  factors. 

Other  methods  for  factoring  will  he  given  in  Chapter  XXIX. 


IX.— GREATEST  COMMON  DIVISOR. 

122.  A  Common  Divisor  or  Measure  of  two  or  more  quan- 
tities is  a  quantity  that  will  divide  each  of  them  without  a 
remainder. 

Hence,  any  factor  common  to  two  or  more  quantities  is  a 
common  divisor  of  those  quantities. 

Also,  when  quantities  are  prime  to  each  other,  they  have  no 
common  measure  greater  than  unity. 


54  ALGEBRA. 

123.  The  Greatest  Common  Divisor  of  two  or  more  quan- 
tities is  the  greatest  quantity  that  will  divide  each  of  them 
without  a  remainder. 

Hence,  the  greatest  common  divisor  of  tivo  or  more  quanti- 
ties is  the  product  of  all  the  prime  factors  common  to  those 
quantities. 

By  the  greatest  of  two  or  more  algebraic  quantities,  it  may 
he  remarked,  is  here  meant  the  highest,  with  reference  to  the 
coefficients  and  exponents  of  the  same  letters. 

In  determining  the  greatest  common  divisor  of  algebraic 
quantities,  it  is  convenient  to  distinguish  three  cases. 

CASE    I. 

124.  When  the  quantities  are  monomials. 
1.    Find  the  greatest  common  divisor  of 

42  a3  b2,  70  a2  b  c,  and  98  a4  b3  d2. 

42  a3b2     =2x3x7  a  a  a     bb 
70  a2  be    =2x5x7  aa        b        c 
98  a4  b3d2  =  2  x  7  X  7  aaaabbb     del 


Hence,  G.  C.  D.  =  2  X  7  a  a  b  =  14  d2  b,  Ans.  (Art.  123).  • 

RULE. 

Resolve  the  quantities  into  their  prime  factors,  and  find  the 
product  of  all  the  factors  common  to  the  several  quantities. 

Note.  Any  literal  factor  forming  a  part  of  the  greatest  common  divisor 
will  take  the  lowest  exponent  with  which  it  occurs  in  either  of  the  given 
quantities. 

EXAMPLES. 

Find  the  greatest  common  divisors  of  the  following : 

2.  as  x2,  7  a*  x,  and  3  a  b2. 

3.  G  c5  d\  3  c3  d5,  and  9  c*  d3. 

4.  18  m  n5,  45  m2  n,  and  72  m8  ri2. 

5.  15  c2  x,  45  c3  x2,  and  60  c4  x8. 


GREATEST   COMMON   DIVISOR.  55 

6.  108  y2  z\  144  f  z\  and  120  if  z5. 

7.  96  a5  b\  120  a3  b5,  and  168  a4  b&. 

8.  51  m4  n,  85  ra3  a-,  and  119  m2  if. 

9.  84  a8  y4  zs,  112  a;4  v/5  z6,  and  154  a7  y6  z\ 

CASE    II. 

125.      When  the  quantities  are  polynomials  which  can  be 
readily  factored  by  inspection. 

1.  Find  the  greatest  common  divisor  of 

5xy3  —  15y3,  x2  +  4 x  —  21,  and mx  —  3m  —  nx  +  3n. 

5xy3  —  15y3  =  5y3(x  —  3) 

x2  +  4  x  -  21  =  (x  +  7)  (a;  -  3) 
mx  —  3  m  —  nx  +  3n  =  (m  —  n)  (x  —  3) 

Hence,  by  Art.  123,  G.  C.  D.  =  x  —  3,  Ans. 

2.  Find  the  greatest  common  divisor  of 

4  x2  -  4  x  +  1,  4  x2  -  1,  and  8  ar5  - 1. 

4a;2-4a;  +  l  =  (2a;-l)  (2  a;  - 1) 
4  a2  - 1  =  (2  x  +  1)  (2  x-1) 
8  a?  -  1  =  (2  x  - 1)  (4  a;2  +  2  x  +  1) 

Hence,  G.  C.  D.  =  2  a:  —  1,  Ans. 

The  rule  in  this  case  is  the  same  as  in  Case  I. 

EXAMPLES. 

Find  the  greatest  common  divisors  of  the  following : 

3.  3  a x2  —  2  a2 x,  a2 x2  —  3  a  b x,  and  5  a2 x3  +  2  ax4  —  3  a3x. 

4.  vi2  +  2  in  n  +  n2,  m2  —  n2,  and  m3  +  n3. 

5.  x*  —  1,  a-5  +  a3,  and  a;4  +  2  a;2  +  1. 

6.  3  a  xif  +  21  ay2,  3  c  a-  +  21  c  -  3  d x  -  21  rZ,  and  a;2-3a-  -  70. 

7.  4  x2  — 12  a  +  9,  4  a;2  -  9,  and  4  m2  »a;-6  m2  n. 

8.  9a2-16,  3a;y  — 4^  +  3a;2  — 4a,  and  27 a;3  — 64. 


56  ALGEBRA. 

9.  xs  —  x,  Xs  +  9  x2  —  10  x,  and  x6  —  x. 

10.  a3  —  8  ft3,  5  a  x  +  2  a  —  10  ft  x  —  4  ft,  and  a2  —  4  a  ft  +  4  ft2. 

11.  ar2  -  a;  -  42,  x2  -  4  a  -  60,  and  x2  +  12  x  +  36. 

12.  8  x3  +  125,  4  aj2  -  25,  and  4  a2  +  20  x  +  25. 

13.  3  a  xG  —  3  a  x5,  a  xs  —  9  a  x2  +  8  a  x,  and  2  a  xh  —  2  a  x. 

14.  12  a  x  -  3  a  +  8  c  x  -  2  c,  64  x3  - 1,  and  16  as*  —  8  a?  +  1. 

CASE    III. 

126.  When  the  quantities  are  polynomials  which  cannot  be 
readily  factored  by  inspection. 

Let  a  and  ft  be  two  expressions,  arranged  in  order  of  powers 
of  some  common  letter ;  and  let  the  exponent  of  the  highest 
power  of  that  letter  in  ft  be  either  equal  to  or  less  than  the 
exponent  of  the  highest  power  of  that  letter  in  a.  Suppose 
that  ft  is  contained  in  a,  p  times  with  a  remainder  c ;  suppose 
that  c  is  contained  in  ft,  q  times  with  a  remainder  d ;  and 
suppose  that  d  is  contained  in  c,  r  times  with  no  remainder. 
The  operation  of  division  may  be  shown  as  follows : 

ft)  a    (p 
p  ft 

e)  b    (q 
1  c 


d)  c    (r 
rd 
0 

We  will  first  show  that  d  is  a  common  divisor  of  a  and  ft. 
From  the  nature  of  subtraction,  the  minuend  equals  the  sub- 
trahend plus  the  remainder  ;  hence, 

a=pb  +  c,  ft  =  q  c  +  d,  and  c  =  rd. 

Substituting  r  d  for  c  in  tho  value  of  ft,  we  have 

b  =  q  r  d  +  d  =  d  (q  r  +  1). 


GREATEST   COMMON   DIVISOR.  57 

Substituting  q  r  d  +  d  for  b,  and  r  d  for  c  in  the  value  of  a, 
we  have 

a=p  q  r  d  +  p  d  +  r  d  =  d  (p  q  r  +  p  +  r). 

Hence,  as  d  is  a  factor  of  a  and  also  of  b,  it  is  a  common 
divisor  of  a  and  b. 

We  will  now  show  that  every  common  divisor  of  a  and  b  is 

a  divisor  of  d.     Let  k  he  any  common  divisor  of  a  and  b,  such 

that  fl^ffli  and  b  =  n  k.     From  the  nature  of  subtraction, 

the   minuend  minus   the   subtrahend    equals    the    remainder ; 

hence, 

c  =  a  —  p  b,  and  d  =  b  —  q  c. 

Substituting  m  k  for  a,  and  n  k  for  b  in  the  value  of  c,  we 

have 

c  =  m  k  —  p  n  k. 

Substituting  mk  —  pnk  for  c,  and  n  k  for  b  in  the  value  of 
d,  we  have 

d  —  nk  —  q  (j)ik—pnk)  =  nk  —  qmk+pqnk 
=  k  (n  —  q  m  +  p  q  n). 

Hence,  k  is  a  factor  or  divisor  of  d. 

Therefore,  since  every  common  divisor  of  a  and  b  is  a  divisor 
of  d,  and  no  expression  greater  (Art.  123)  than  d  can  be  a 
divisor  of  d,  it  follows  that  d  is  the  greatest  common  divisor 
of  a  and  b. 

1.  Find  the  greatest  common  divisor  of  x1  —  6  x  +  8  and 
4  x3  -  21  x2  +  15  x  +  20. 

x2  -  6  x  +  8)  4  a3  -  21  x2  +  15  x  +  20  (4  x  +  3 

4  x3  -  24  a;2  +  32  x     

3  x2  - 17  x  +  20 
3  a;2  -  18  a;  +  24 

x  _  4)  ^  —  6x  +  8(x-2 
x2  —  4aj 
.      -2a;  +  8 
-2a  +  8 

Hence,  cc  —  4  is  the  greatest  common  divisor,  Ans. 


58  ALGEBRA. 

RULE. 

Divide  the  greater  quantity  (Art.  123)  by  the  less  •  and  if 
there  is  no  remainder,  the  less  quantity  trill  he  the  required 
greatest  common  divisor. 

If  there  is  a  remainder,  divide  the  divisor  by  it,  and  continue 
thus  to  make  the  preceding  divisor  the  dividend,  and  the  re- 
mainder the  divisor,  until  a  divisor  is  obtained  which  leaves  no 
remainder  ;  the  last  divisor  will  be  the  greatest  common  divisor 
required. 

Nota  1.  If  there  are  three  or  more  quantities,  find  the  greatest  common 
divisor  of  two  of  them  ;  then  of  this  result  and  the  third  of  the  quantities, 
and  so  on.     The  last  divisor  will  be  the  greatest  common  divisor  required. 

Note  2.  If  a  monomial  factor  is  seen  by  inspection  to  be  common  to  all 
the  terms  of  one  of  the  given  quantities,  and  not  of  the  other,  it  may  be  re- 
moved, as  it  evidently  can  form  no  part  of  the  greatest  common  divisor ; 
and,  similarly,  we  may  remove  from  a  remainder  any  monomial  factor 
which  is  not  a  common  factor  of  the  given  quantities. 

2.    Find  the  greatest  common  divisor  of 

6  a  x~  —  19  a  x  +  10  a  and  6  xs  —  x2  —  35  x. 

In  the  first  quantity  a  is  a  common  factor  of  all  the  terms, 
and  is  not  a  factor  of  the  second  quantity  ;  in  the  second  quan- 
tity x  is  a  common  factor  of  all  the  terms,  and  is  not  a  factor 
of  the  first  quantity.  Hence  we  may  remove  a  from  each 
term  of  the  first  quantity,  and  x  from  each  term  of  the  second. 

6  a;2- 19  a  +  10)6  cc2-     x  -35(1 

6a:2-- 19:c  + 10 
18a;-45 

In  this  remainder  9  is  a  common  factor  of  all  the  terms,  and 
is  not  a  common  factor  of  the  given  quantities.  Hence  9  may 
be  removed  from  each  term  of  the  remainder. 

2  x  -  5)6  x1  -  19  x  +  10(3  x  -  2 
6  x2  —  15  x 

—  4  x'+  10 

—  4  a; +  10 

Hence,  2  x  —  5  is  the  greatest  common  divisor,  Ans. 


GREATEST   COMMON   DIVISOR.  59 

Note  3.     If  the  first  term  of  a  remainder  be  negative,  the  sign  of  each 
term  may  be  changed. 

3.    Find  the  greatest  common  divisor  of  2  x2  —  3  a;  —  2  and 
2a;2-5a;-3. 

2a;2  -  3  x  -  2)2  a2-  5  x  -3(1 
2x2-3x-2 


-2x-l 

The  first  term  of  this  remainder  being  negative,  we  change 
the  sign  of  each  term,  giving  2  x  +  1. 

2  x  +  1)2  a;'2-  3  x  -  2 (x  —  2 
2  x'2  +     x 


—  4x  —  2 

—  4a;  — 2 

Hence,  2  x  +  1  is  the  greatest  common  divisor,  Ans. 

Note  4.  The  dividend  or  any  remainder  may  be  multiplied  by  any 
quantity  which  is  not  a  common  factor  of  all  the  terms  of  the  divisor. 

4.  Find  the  greatest  common  divisor  of  2  a-3  —  7  xr  +  5  x  —  6 
and  3  a'3  —  7  a'2  —  7  x  +  3. 

To  avoid  a  fraction  as  the  first  term  of  the  quotient,  we 
multiply  each  term  of  the  second  quantity  by  2,  giving 
6  Xs  -  14  x2  -  14  x  +  6. 

2  xs—  7  x2  +  5  x  —  6)6  xs  - 14  cc2  - 14  x  +    6  (3 

6  a;3  -  21  x2  +  15  a;  -  18 


7  a-2 -29  a; +  24 


To  avoid  a  fraction  as  the  first  term  of  the  next  quotient, 
we  multiply  each  term  of  the  new  dividend  by  7,  giving 
14  x3  -  49  x2  +  35  x  -  42. 

7  a;2  -  29  x  +  24)  14  x3  -  49  x2  +  35  x  -  42  (2  a; 
14  a;3  -  58  a2  +  48  x 

9  a-2  - 13  x  -  42 


60  ALGEBRA. 

The  first  term  of  this  remainder  not  heing  exactly  divisible 
by  the  first  term  of  the  divisor,  we  multiply  each  term  hy  7, 
giving  63  x2  —  91  x  —  294. 

7  x2 - 29 x  +  24) 63 x2-   91^-294(9 
63  x2  -  261  x  +  216 


170  x  -  510 

Dividing  each  term  by  170,  x  —  3)  7  x2  —  29  x  +  24  (7  x  —  8 

7  x2  -21x 


—  8  a; +  24 

-  8cc  +  24 

Hence,  x  —  3  is  the  greatest  common  divisor,  Ans. 

Note  5.  When  the  two  given  quantities  have  a  common  monomial 
factor,  it  may  be  removed  from  each,  and  the  greatest  common  divisor  of 
the  resulting  expressions  found.  This  result  must  be  multiplied  by  the 
common  monomial  factor  to  give  the  greatest  common  divisor  of  the  given 
quantities. 

5.    Find  the  greatest  common  divisor  of  6  x3  —  x2  —  5  x  and 

21  x3  -  26  x2  +  5x. 

The  quantities  have  the  common  monomial  factor  x ;  remov- 
ing it,  we  find  the  greatest  common  divisor  of  6  x2  —  x  —  5  and 
21  x2  —  26  x  +  5.  We  multiply  the  latter  by  2,  to  avoid  a  frac- 
tion as  the  first  term  of  the  quotient,  giving  42  a;2  —  52  x  +  10. 

6  x2  -  x  -  5)  42  x2  -  52  x  +  10  ( 7 
42  x2-    7a;-35 


—  45  x  +  45 


Dividing  by  —  45,     x  —  1)6  x2—     x  —  5(6a:  +  5 

6  x2  —  6  x 


5x  —  5 
5x  —  5 


LEAST   COMMON   MULTIPLE.  61 

Hence,  x  —  1  is  the  greatest  common  divisor  of  6  x2  —  x  —  5 
and  21  x2  —  26  x  +  5.  Multiplying  by  x,  the  common  mo- 
nomial factor,  we  obtain  x  (x  —  1)  or  x2  —  x  as  the  required 
greatest  common  divisor,  Ans. 

EXAMPLES. 

Find  the  greatest  common  divisors  of  the  following : 

6.  6  x2  —  1  x  —  24  and  12  x2  +  8  x  -  15. 

7.  24  x2  +  11  x  -  28"  and  40  x2  -  51  x  +  14. 

8.  2  xs -2  x2 -  3 x  +  3  and  2 x3 -2  x2-  ox  +  5. 

9.  6 x2- 13 x -28  and  15 x2  +  23 a;  +  4. 

10.  8  x2-  22  x  +  5  and  6  a;2  -  23  x  +  20. 

11.  5  x2  +  58  jc  +  33  and  10  .x2  +  41  x  +  21. 

12.  x3  +  2  x2  +  x  +  2  and  xi  -4r-x-2. 

13.  2a;3-3cc2-;c  +  l  and  6xs  —  x2  +  3x -2. 

14.  a,-4  -  x3  +  2  x2  +  x  +  3  and  xi  +  2  x3  —  x  —  2. 

15.  ft2-5ax  +  4 x2  and  a3  —  «2 x  +  3  a x2  —  3 x3. 

16.  xA  -x3-5  x2  +  2  a-  +  6  and  x4  +  x3  -  x2  -2x-  2. 

17.  6  x2  y  +  4  x  y2  —  2  y3  and  4  cc3  +  2  x2  ?/  —  2  a*  //\ 

18.  2«4  +  3a3x-9a2x2and  6  «3- 17  a2;c  +  14  a  ar- 3x3. 

19.  15  a2  x3  -  20  a2  x2  -  G5  a2  x  -  30  a2  and  12  6  a;3  +  20  b  x2 

— 16  b  x  — 16  b. 


X.  — LEAST  COMMON  MULTIPLE. 

127.     A  Multiple  of  a  quantity  is  any  quantity  that  can  be 
divided  by  it  without  a  remainder. 

Hence,  a  multiple  of  a  quantity  must  contain  all  the  prime 
factors  of  that  quantity. 


62  ALGEBRA. 

128.  A  Common  Multiple  of  two  or  more  quantities  is  one 
that  can  be  divided  by  each  of  them  without  a  remainder. 

Hence,  a  common  multiple  of  two  or  more  quantities  must 
contain  all  the  prime  factors  of  each  of  the  quantities. 

129.  The  Least  Common  Multiple  of  two  or  more  quanti- 
ties is  the  least  quantity  that  can  be  divided  by  each  of  them 
without  a  remainder. 

Hence,  the  least  common  multiple  of  two  or  more  quantities 
must  be  the  pmxluct  of  all  their  different  prime  factors,  each 
taken  only  the  greatest  number  of  times  it  is  found  in  any  one 
of  those  quantities. 

By  the  least  quantity,  is  here  meant  the  lowest  with  refer- 
ence to  the  exponents  and  coefficients  of  the  same  letters. 

In  determining  the  least  common  multiple  of  algebraic 
quantities,  we  may  distinguish  three  cases. 

CASE    I. 

130.  When  the  quantities  are  monomials. 

1.  Find  the  least  common  multiple  of  36  as  x,  60  a2  y2,  and 
84  c  xs. 

36  a3  #  =  2x2x3x3  a  a  ax 
60aV=2x2x3x5  a  a  yy 

84  c  xz  =2x2x3x7  x  x  x         c 

Hence,  L.  C.  M.  =  2  xf2  X3x3x5  X?  aaaxxxyyc 
=  1260  a3  xs  if  c,  Ans.  (Art.  129). 

RULE. 

Resolve  the  quantities  into  their  prime  factors;  and  the 
product  of  these,  taking  each  factor  only  the  greatest  number  of 
times  it  enters  into  any  one  of  the  quantities,  will  be  the  least 
common  multiple. 

Any  literal  factor  forming  a  part  of  the  least  common  mul- 
tiple will  take  the  highest  exponent  with  which  it  occurs  in 
p.ither  of  the  given  quantities. 


LEAST   COMMON    MULTIPLE.  6 


o 


When  quantities  are  prime  to  each  other,  their  product  is 
their  least  common  multiple. 

EXAMPLES. 

Find  the  least  common  multiples  of  the  following : 

2.  8  a4  c,  10  a3  b,  and  12  a2  b2. 

3.  5  x3  y,  10  if  z,  and  15  x  z3. 

4.  a5  b2,  9  a3  b\  and  12  a2  b3. 

5.  24  m3  x2,  30  n2  y,  and  32  x  y2. 

6.  8  c2  d3,  10  a  e,  and  42  a2  d. 

7.  36  x  y2  z3,  63  x3  y  z2,  and  28  .r2  y3  z. 

8.  40  a2  b  d3,  18  a  c3  d\  and  54  J2  c  d\ 

9.  7  m  w2,  8  x3  y2,  and  84  n  x  y3. 

CASE    II. 

131.     When  the  quantities  are  polynomials  which  can  be 
readily  factored  by  inspection. 

1.    Find  the  least  common  multiple  of  x2  +  x  —  6,  x2—  6  x  +  8 
and  x2  —  9. 

x*  +     a,  _  6  =  (x  -  2)  (x  +  3) 

a;2_6a:  +  8  =  (a;-2)  (cc-4) 
x2-9  =(x-3)(x  +  3) 

Hence  (Art.  129),  L.  C.  M.  =  (x  -  2)  (a;  -  3)  (sc  +  3)  (x  -  4) 
or,  x4  -  6  x3  -  x2  +  54  x  -  72,  ^ws. 

The  rule  is  the  same  as  in  Case  I. 

EXAMPLES. 

Find  the  least  common  multiples  of  the  following : 

2.  a  x2  +  a2  x,  x2  —  a2,  and  x3  —  a3. 

3.  2  a2  +  2  a  b  ,  3  a  b  -  3  &2,  and  4  a2  c  —  4  ft2  c. 


64  ALGEBRA. 

4.  x2  +  x,  xz  —  x,  and  xi  +  x. 

5.  2  -  2  a-2,  4  -  4  a-,  8  +  8  a,  and  12  +  12  a2. 

6.  x2+  5x  +  4,  x2  +  2x  —  S,  and  a;2  +  7  jc  +  12. 

7.  x3  —  10  a;2  +  21  a;,  and  a  x2  +  5  a  x  —  24  a. 

8.  4  ar  -  4  a;  +  1,  4  x2  -  1,  and  8  a3  -  1. 

9.  a  x  —  a  y  —  b  x  +  b  y,  x2  —  2  x  y  +  y2,  and  3  arb  —  3ab2. 

10.  9  a;2  +  12  a;  +  4,  27  x3  +  8,  and  6  a  x3  +  4  a  x2. 

11.  cc2  -  4  cc  +  3,  x2  +  a;  -  12,  and  x2-x-  20. 

12.  x2  —  y2  —  z2  +  2  y  z  and  x2-i/2+r  +  2a;s. 

CASE    III. 

132.  When  the  quantities  are  polynomials  which  cannot  be 
readily  factored  by  inspection. 

Let  a  and  b  be  two  expressions ;  let  d  be  their  greatest  com- 
mon divisor,  and  m  their  least  common  multiple.  Suppose 
that  d  is  contained  in  a,  x  times,  and  in  b,  y  times ;  then,  from 
the  nature  of  the  greatest  common  divisor,  x  and  y  are  prime 
to  each  other.  Since  the  dividend  is  the  product  of  the  quo- 
tient and  divisor,  we  have 

a  =  dx  and  b  =  d  y. 

Then  (Art.  129)  the  least  common  multiple  of  a  and  b  is 

d  x  y,  or  m  =  d  x  y  ;  but  dx  =  a,  and  y  =  -',   substituting,  we 

h 
have  m  =  a  X  -;  • 

d 

In  a  similar  manner  we  could  show  that  m  —  b  X  -;• 

d 

Hence  the  following 

RULE. 

Find  the  greatest  common  divisor  of  the  two  quantities  ;  di- 
vide one  of  the  quantities  by  this,  and  multiply  the  quotient  by 
the  other  quantity. 


LEAST   COMMON   MULTIPLE.  65 

Note.  If  there  are  three  or  more  quantities,  find  the  least  common 
multiple  of  two  of  them,  and  then  of  that  result  and  the  third  quantity  ; 
and  so  on. 

1.  Find  the  least  common  multiple  of  6  x2  —  17  x  +  12  and 
12  a:2- 4  a -21. 

6  a;2- 17  x  + 12)12  a:2-   4sc-21(2 
12  x2  -  34  x  +  24 


30  x  -  45 
2  x  -  3 )  6  x2  - 1 7  x  +  1 2  ( 3  x  -  4 
6  x2  —   9  x 


-  8  x  +  12 

-  8  a  +  12 


Hence,  2  a:  —  3  is  the  greatest  common  divisor  of  the  two 
quantities  ;  dividing  the  first  given  quantity  by  this,  we  obtain, 
as  a  quotient,  3  x  —  4  ;  multiplying  the  second  given  quantity 
by  this  quotient,  we  have 

(3  a; -4)  (12  a;2 -4  a: -21),  or  36  x3  -  60  x2  -  47  x  +  84 
as  the  required  least  common  multiple,  Ans. 

EXAMPLES. 

Find  the  least  common  multiples  of  the  following : 

2.  6  x2  +  13  x  -  28  and  12  x2  -  31  x  +  20. 

3.  8  x2  +  30  x  +  7  and  12  x2  -  29  x  -  8. 

4.  a3  +  a2_  8  a  _  6  and  2  a3  -  5  a2-  2  a  +  2. 

5.  2  x3  +  x2  -  x  +  3  and  2  .t3  +  5  x2  -  x  -  6. 

6.  (t3-2«2H2ffii2-  63  and  a3  +  a2  6  -  a  b2  -  b3. 

7.  x*  +  2  x3  +  2  x2  +  x  and  a  «3  —  2  a  x  —  a. 

8.  2x4-llx-3+3a;2  +  10a;  and  3a;4- 14z3- 6ar+ 5oj. 


66  ALGEBRA. 


XL  — FRACTIONS. 

133.  A  Fraction  is  an  expression  indicating  a  certain 
number  of  the  equal  parts  into  which  a  unit  has  been  divided. 

The  denominator  of  a  fraction  shows  into  how  many  parts 
the  unit  has  been  divided,  and  the  numerator  how  many  parts 
are  taken. 

134.  A  fraction  is  expressed  by  writing  the  numerator 
above,  and  the  denominator  below,  a  horizontal  line.  Thus, 
-  is  a  fraction,  signifying  that  the  unit  has  been  divided  into 

b  equal  parts,  and  that  a  parts  are  taken. 

The  numerator  and  denominator  are  called  the  terms  of  a 
fraction. 

Every  integer  may  be  considered  as  a  fraction  whose  denomi- 

a 
nator  is  unity  ;  thus,  a  =  r- . 

135.  An  Entire  Quantity  is  one  which  has  no  fractional 
part ;  as,  ab,  or  a  —  b. 

136.  A  Mixed  Quantity  is  one   having  both  entire  and 

b  a 

fractional  parts  ;  as,  a ,  or  c  + 


x  +  y 

137.  If  the  numerator  of  a  fraction  be  multiplied,  or  the 
de nominator  divided,  by  any  quantity,  the  fraction  is  multi- 
plied by  that  quantity. 

1.    Let  y  denote  any  fraction  ;  multiplying  its  numerator  by 

c,  we  have  -— .     Now,  in  -  and  —  the  unit  is  divided   into  b 
b  b  b 

equal  parts,  and  a  and  a  c  parts,  respectively,  arc  taken.     Since 


FRACTIONS.  67 


c  times  as  many  parts  are  taken  in  —  as  in  - ,  it  follows  that 

a  c  .       .,.         a 

— -  is  c  times  -. 

b  o 

2.    Let  —  denote   any  fraction  ;    dividing   its   denominator 

a  a  a 

by  c,  we  have  -.     Now,  in  —  and  -,    the    same    number   of 

b  be    a    b 

parts  is  taken  ;    but,   since   in  —    tbe    unit    is    divided    into 

a  .  hc 

b  c  equal  parts,  and  in  -  into  only  b  equal  parts,  it  follows  that 

•     a  .         .  ,  i  ,  •      a      tt  a 

each  part  m  -  is  c  times  as  large  as  each  part  m  — .    Hence,  - 

is  c  times  -r—  . 
be 

138.  If  the  numerator  of  a  fraction  be  divided,  or  the  de- 
nominator multiplied,  by  any  quantity,  the  fraction  is  divided 
by  that  quantity. 

1.  Let  —  denote  any  fraction  ;  dividing  its  numerator  by  c, 

we  have-.     Now,  in  Art.  137,  1,  we  showed  that  —  was  c 
a  o  a  .    ac     .   .  b 

times  -.     Hence,  -  is  —  divided  by  c. 
b  b      .  b 

2.  Let  -  denote  any  fraction :  multiplying  its  denominator 

b 

a  a 

by  c,  we  have  — .     Now,  in  Art.  137,  2,  we  showed  that  -  was 

a         bc  a        a    .   .  h 

e  times  — .     Hence,  —  is  -  divided  by  c. 
b  c  b  c       b 

139.  If  the  terms  of  a  fraction  be  both  multiplied,  or  both 
divided  by  the  same  quantity,  the  value  of  the  fraction  is  not 

altered. 

For,  multiplying  the  numerator  by  any  quantity,  multiplies 
the  fraction  by  that  quantity ;  and  multiplying  the  denomi- 
nator by  the  same  quantity,  divides  the  fraction  by  that 
quantity.  And,  by  Art.  44,  Ax.  6,  if  any  quantity  be  both 
multiplied  and  divided  by  the  same  quantity,  its  value  is  not 
altered. 


68  ALGEBRA. 

Similarly,  we  may  show  that  if  Loth  terms  are  divided  by 
the  same  quantity,  the  value  of  the  fraction  is  not  altered. 

140.  We  may  now  show  the  propriety  of  the  use  of  the 
fractional  form  to  indicate  division,  as  explained  in  Art.  16  ; 

ft 

that  is,  we  shall  show  that  -  represents  the  quotient  of  a  di- 
vided by  b. 

For,  let  x  denote  the  quotient  of  a  divided  by  b. 

Then,  since  the  quotient,  multiplied  by  the  divisor,  gives 
tbe  dividend,  we  have  b  x  =  a. 

But,  by  Art.  137,  bXj=a. 

Therefore,  x  =  - . 

b 

141.  A  fraction  is  positive  when  its  numerator  and  de- 
nominator have  the  same  sign,  and  negative  when  they  have 
different  signs. 

For,  a  fraction  represents  the  quotient  of  its  numerator 
divided  by  its  denominator ;  consequently  its  proper  sign  can 
be  determined  as  in  division  (Art.  91). 

142.  The  Sign  of  a  fraction  is  the  sign  prefixed  to  its 
dividing  line,  and  indicates  whether  the  fraction  is  to  be 
added  or  subtracted. 

Thus,  in  x  -\ — —  the  sign  +  denotes  that  the  fraction  -j— , 

although  essentially  negative  (Art.  91),  is  to  be  added  to  x. 

The  sign  written  before  the  dividing  line  of  a  fraction  is 
termed  the  apparent  sign  of  the  fraction  ;  and  that  de] tending 
upon  the  value  of  the  fraction  itself  is  termed  the  real  sign. 

Thus,  in  -\ — — ,  the  apparent  sign  is  + ,  but  the  real  sign 
is  — . 

Where  no  signs  are  prefixed,  plus  is  understood. 


ah 

—  ah           ah 

-ah 

h 

b              -b 

-b 

ah 

a  b      —  ah 

—  ah 

— _  i 





b 

-b         b 

-b 

FRACTIONS.  69 

143.  From  the  principles  of  Arts.  140  and  141  we  obtain, 

+  a; 

—  a. 
—  o 

From  which  it  appears  that, 

Of  the  three  signs  prefixed  to  the  numerator,  denominator, 
and  dividing  line  of  a  fraction,  any  two  may  he  changed  with- 
out altering  the  value  of  the  fraction  ;  hut  if  any  one,  or  all 
three  are  changed,  the  value  of  the  fraction  is  changed  from 
+  to  — ,  or  from  —  to  +  . 

144.  If  either  the  numerator  or  denominator  of  the  frac- 
tion is  a  polynomial,  we  mean  by  its  sign  the  sign  of  the  entire 
expression,  as  distinguished  from  the  sign  of  any  one  of  its 
individual  terms  •  and  care  must  be  taken,  pn  changing  signs 
in  any  such  case,  to  change  the  sign  before  each  term. 

„,,  a  —  h—a  +  b         b  —  a 

Thus'         -c^r-c^d>ovc—}v 

a—h—a+b        h  —  a 
also,  j  = -.,  or . 

c  —  d      —c  +  d         d  —  c 

145.  From  Art.  141  we  have 

abed  _(-a)h  (-  c)  (—d)  __  a  (-  b)  (-  c)  d 


efffh  (~e)fgh  e(-f)g(-h) 


,  etc. ; 


abed  _  (—  a)bc{—  d)  _       a  (—  h)  (—  c)  d 
~  Jffh  -     (re)fgY  ~  e  (-/)  (-  g)  (-  h)'  : 

From  which  it  appears  that, 

If  the  terms  of  a  fraction  are  composed  of  any  number  of 
factors,  any  even  number  of  factors  may  have  their  signs 
changed  without  altering  the  value of  the  fraction  •  but  if  any 


70  ALGEBRA. 

odd  number  of  factors  have  their  signs  changed,  the  value  of 
the  fraction  is  changed  from  +  to  — ,  or  from  —  to  +. 

a  —  b  a  —  b  b  —  a 


Thus, 


(x  —  y)(x  —  z)~    (y  —  x)  (z  —  x)~    {y^-  x)  {x  —  z) 


b  —  a  1        n  .  b  —  a 

but  does  not  equal 


(x  —  y)(z  —  x)  (y  —  x){z  —  x) 


REDUCTION   OF   FRACTIONS. 

146.  Reduction  of  Fractions  is  the  process  of  changing 
their  forms  without  altering  their  values. 

TO  REDUCE  A  FRACTION  TO  ITS  SIMPLEST  FORM. 

147.  A  fraction  is  in  its  simplest  form,  when  its  terms  are 
prime  to  each  other. 

CASE    I. 

148.  When  the  numerator  and  denominator  can  be  readily 
factored  by  inspection. 

Since  dividing  hoth   numerator   and   denominator  by  the 

same  quantity,  or  cancelling  equal  factors  in  each,  does  not 

alter  the  value  of  the  fraction   (Art.  139),  we  have  the  fol- 
lowing 

RULE. 

Resolve  both  numerator  and  denominator  into  their  prime 
factors,  and  cancel  all  that  are  common  to  both. 

EXAMPLES. 

,    ^,   ,        18  a8  b2  o       .       .      ,       , 

1.  ixeduce  — — ^—2 —  ™  ^s  simplest  form. 

18  a*b2  c  __  2 .  3  . 3 .  a .  a .  a .  b .  h .  c     2ac 
45  a2  b'2  x  ~  5 .  3 . 3 .  a  .  a .  b  .  b .  x  5  x  '  * 

x2+  2  x  — 15 

2.  Eeduce  — s — —  to  its  simplest  form. 

x-  —  2x  —  3 

g;2+  2x -15_  (x  +  5)  (x - 3)  _ ,-r  +  5 
x2  -  2  x  -  3  : "  (x  +  1)  (x  -  3)  ~  x~+l' 


FRACTIONS.  71 

_     _.    ,  b  c,  —  a  r  —  b  d  +  a  d         .        .       . 

3.    Eeduce : ; —  to  its  simplest  form. 

a  in  —  b  in  —  an  +  0  n 

be  —  ac  —  bd  +  ad        (b  —  a)  (c  —  d) 

am  —  b  m  —  an+  bn      (a  —  b)  (m  —  n) 

=  (Art.  89)  ("-f)(<*-«0  =  fLllt  Ans. 
(it  —  0)  (m  —  n)      m  —  n 

Note.  If  all  the  factors  of  the  numerator  be  removed  by  cancellation, 
the  number  1  (being  a  factor  of  all  algebraic  expressions)  remains  to  form  a 
numerator. 

If  all  the  factors  of  the  denominator  be  removed,  the  result  will  be  au 
entire  quantity  ;   this  being  a  case  of  exact  division. 

Reduce  the  following  to  their  simplest  forms : 
4.  SfUU.  13. 


c  K.9U    !IV    IV  -    - 

oo  mr  n6 

65x2y3z* 
2(>  x*  y°  z~ 

„    54  a3  b5  c2 
72  a2  b2  c 

Wmxif  ,« 

o.    — —  •  1 1 . 

to  m  x  y- 

110  e3  x2  y 
9-      22c2x2    ■  1S' 

1A    2a2cd+2abcd  1Q 

1U-    a~^ Tr — 7 '         iy> 

b  <r  x  y  +  b  ab,x  y 

ii.     3»'-e«4y  go. 

6x2y2  —  12  xy3  ac  +  ad  —  b  c  —  b  d 

19     x2  —  2x  —  lh'  2mx  + 3my  —  2n2x  —  3n2y 

x2  +  10 x  +  21  '  2 m2 x  +  3  m2 y—2nx—3n  y 


m2 —  10m 

+  16 

m2  +  m  - 

-72    * 

4  c2  -20c 

'  +  25 

25-4 

c2       ' 

4  a  — 

9  a  n2 

9bn2-12bn  +  4:b 

8  x3  +  y3 
4  x2  —  y2 ' 

27  y3  -  125 


25- 

-30  7/  +  9y2  ' 

6  a;2 

y  —  2  x3  y 

x2- 

-8x  +  15  " 

4  —  x2 

X3- 

-  9  x2  +  14  x  ' 

a  e 

—  b  c  —  ad +  bd 

72  ALGEBRA. 

CASE     II. 

149.  When  the  numerator  and  denominator  cannot  be 
readily  factored  by  inspection. 

Since  the  greatest  common  divisor  of  two  quantities  con- 
tains all  the  prime  factors  common  to  both,  we  have  the  fol- 
lowing 

KULE. 

Divide  both  numerator  and  denominator  by  their  greatest 
common  divisor. 

EXAMPLES. 

1.   Reduce  — - — 5 — —  to  its  simplest  form. 

6  cr  —  a  —  12 

By  the  rule  of  Art.  126,  we  find  the  greatest  common  divisor 
of  the  numerator  and  denominator  to  he  2.a  —  3.  Dividing 
the  numerator  by  this,  the  quotient  is  a  —  1.  Dividing  the 
denominator,  the  quotient  is  3  a  +  4.     Therefore,  the  simplest 

form  of  the  fraction  is -,  Ans. 

3  a  +  4' 

Reduce  the  following  to  their  simplest  forms : 

6a:2  +  a:-35  „     6  a-3  -  19  x2  +  7  x  +  12 


o      -«-"  "    —  "-  —  "J-  g 

'  '2u2-7a  +  6' 

.        2  m2  —  5  m  +  3  Q 

'    12  m3  -  28  m  +  15  ' 

xa  +  x*-Sx-2 

O.      -= ; s ~ 7Z  •  1U. 


6'    2x3  +  5x2-2:c  +  3" 


8x2  + 

22  a: +  5' 

10  a2- 

-a-21 

2u2- 

7  a  +  6' 

2  m2 

—  5  m  + 

3 

12  m2- 

-  28  m  + 

15  ' 

xs  +  x2  —  3  x  - 

-2 

x3  — 4 

x2  +  2x- 

4-3" 

6  a3  — 

7  x2  +  5  a 

•-2 

6a;3-25a:2  +  17a-  + 

20 

4a:3  +  14ar+12a-  + 

5 

4a;3-10a;2-12a'- 

rr' 

1 

12  a2  +  16  a  -  3 

10  a2  +  a  -  21  ' 

x3  —  4  x2  +  4  a;  —  1 

a3  -  2  x2  +  4  aj  -  3  " 

6  x3  —  x2  —  7  x  —  2 

FRACTIONS.  73 

TO  REDUCE  A  FRACTION  TO  AN  ENTIRE  OR  MIXED  QUANTITY. 

150.     Since  a  fraction  is  an   expression   of  division  (Art. 
140),  we  have  the  following 

RULE. 

* 

Divide  the  numerator  by  the  denominator,  and  the  quotient 
will  be  the  entire  or  mixed  quantity  required. 


EXAMPLES. 


ax  —  a2  x2 


1.  Reduce   —  to  an  entire  quantity. 

(ax  — a2 x2)  -i-ax  =  l  —  ax,  Ans. 

q%  A3 /j.3 

2.  Reduce  to  a  mixed  quantity. 

b3 

a  —  x)a3  —  x3  —  b3(a2  +  a  x  +  x2 ,  Ans. 

a  —  x 


a3  —  a2  x 

a2  x  — 

x3  — 

b3 

9 

a  x  — 

a  x2 

ax2 

—  X3 

— 

bz 

ax2 

—  x3 

-b3 

Reduce  the  following  to  entire  or  mixed  quantities : 

ab  —  a2  o~2  ■    k 

o. 

4. 

5. 

6. 

7. 

2ab 


b        ' 

X3 

+  y3 

X 

+  y 

2: 

K2-3x-4: 

5  x 

X3 

2     i     rr 

—  x'  +  7x  — 

6 

3  x 

a2 

—  3ab  +  4b2 

8. 

x  —  3 

9. 

Xs— 1 
X-V 

10. 

4,x2-2x  +  5 

2x2-x  +  \  ' 

11 

/Y»<*   ___    ryi"—   /y»    i     i  _     */ 

*As                  *Aj                 *as               *J 

X2  +  X  —  1 

12 

2cc3-3.x2+4r-2 

2  x2  —  3  x  +  3 

74  ALGEBRA. 

TO  REDUCE  A  MIXED  QUANTITY  TO  A  FRACTIONAL  FORM. 

151.  This  is  the  converse  of  Art.  150;  hence  we  may- 
proceed  by  the  following 

RULE. 

Multiply  the  entire  part  by  the  denominator  of  the  fraction  ; 
add  the  numerator  to  the  product  when  the  sign  of  the  fraction 
is  + ,  and  subtract  it  when  the  sign  is  —  /  writing  the  result 
over  the  denominator. 

EXAMPLES. 

a2 £2 5 

1.  Eecluce    a  +  b — : —  to  a  fractional  form. 

a  —  b 

By  the  rule, 

a2  _  b2  _  5       (a  +  &)  (g-b)-  Q2  -  b°-  -  5) 


a  —  b  a  —  b 


a  —  b  a  —  b 

Note.     It  will  be  found  convenient  to  enclose  the  numerator  in  a  pa- 
renthesis, when  the  sign  before  the  fraction  is  —  . 

Reduce  the  following  to  fractional  forms: 

4  „    0         3a2-2Z>2 

2.  x  +  l  + .-  7.2a ^- 

x  —  6  a  a 

3.  a  +  — 8.    a2+ab  +  b2—7 • 

n  b  —  a 

4.  7a;-4"2  +  5a-  9.  3z-2-      3 


~8  2x-l 

i      x  +  1  in  /      a*  +  b* 

5.  *  +  i  +  __.  10.   a-&_-^-. 

a  +■  «  a;  — ^s 


FRACTIONS.  75 

TO  REDUCE  FRACTIONS  TO  A  COMMON  DENOMINATOR. 

t  co     1     x>    i  5cd     3mx        j  3  n  y 

152.     1.    Keduce    - — — , ,  and  — -^     to    a   common 

3  crb     2  ab2  ka6b 

denominator. 

Since  multiplying  each  term  of  a  fraction  by  the  same  quan- 
tity does  not  alter  the  value  of  the  fraction  (Art.  139),  we 
may  multiply  each  term  of  the  first  fraction  by  4  a  b,  giving 

20  a  b  c  d  ,  «  ,  1  ■,  -.      0    „      .  .        18  a2  m  x 

;  each  term  of  the  second  by  b  a  ,  giving  ; 

12  a?  62     '  J  '  &        °     12«:i62 

and  each  term  of  the  third  by  3  b,  giving  L. . 

12  aJ  b" 

It  will  be  observed  that  the  common  denominator  is  the 
least  common  multiple  of  the  given  denominators,  which  is 
also  called  the  least  common  denominator  ;  and  that  each  term 
of  either  fraction  is  multiplied  by  a  quantity  which  is  obtained 
by  dividing  the  least  common  denominator  by  its  own  denomi- 
nator.    Kence  the  following 

RULE. 

Find  the  least  common  multiple  of  the  given  denominators. 
Divide  this  by  each  denominator,  separately,  and  multiply  the 
corresponding  numerators  by  the  quotients  ;  writing  the  results 
over  the  common  denominator. 

Before  applying  the  rule,  each  fraction  should  be  in  its  sim- 
plest form ;  entire  and  mixed  quantities  should  be  changed  to 
a  fractional  form  (Arts.  134  and  151). 

Note.  The  common  denominator  may  be  any  common  multiple  of  the 
given  denominators.  The  product  of  all  the  denominators  is  evidently  s 
common  multiple,  and  the  rule  is  sometimes  given  as  follows  :  "Multiply 
'each  numerator  by  all  the  denominators  except  its  own,  and  write  the 
results  over  the  product  of  all  the  denominators." 

an  a  x  x  it 

2.    Reduce  - — —  ,  — — ,  and    . .  ■       N,  to  a  common  de- 

1  —  x     (1  —  x)2  (1  —  xy 

nominator. 


76  ALGEBRA. 

The  least  common  multiple  of  the  given  denominators  is 
(1  —  be)3.  Dividing  this  by  the  first  denominator,  the  quotient 
is  (1  —  a:)2;  dividing  it  by  the  second  denominator,  the  quo- 
tient is  (1  —  x)  ;  and  dividing  it  by  the  third  denominator,  the 
quotient  is  1.  Multiplying  the  corresponding  numerators  by 
these  quotients,  we  obtain  a  y  (1  —  x)2,  a  ar  (1  —  x),  and  x  i/3 
as  the  new  numerators.     Hence  the  results  are 

a  y  (1  —  a?)2    ax2  (1  —  x)  x  y3 

(i-xy  >    {i-xy  > and  (i  -  xy ' Ans- 

EXAMPLES. 

Reduce  the  following  fractions  to  a  common  denominator : 

a    3ab    2ac        -,56c  _    4c  — 1    3b  —  2         1    5a 

3-    -^o—  >  -n— >  and  "To--  6-      o  „  ,    >     K  „  .    ,  and 


8    '     9    '  12   '  3ab  '    5ac    '         6b  c 

.     x2 y   xy z  7 y z2  „       2         3  4 

^**     ~t ,  *  i     t~^    i  and     ^"tt — .  /.    — - — — ,  — -,  and  — -. 
10  '     15   '             30  a3  a;2 '  a  a:3'  at 


a2x 


3y  z    Axz  5a;?/  5  az    3bx  ley  —  m 

b'   Yx-'Yy-'  and  TT "  8-   6^'  87i'  and    10*  s2 

9. -,  — — -,  and  — 


a  —  b  '  a  +  b  '  a2  +  62  ° 

10  #  +  3  a?  +  1  a;  +  2 

a;2- 3  a: +  2'  a;2-  5x  +  6'  x2-  4a;  +  3' 

2a  3b  4c 

cr  +  a  —  6 '  a2+  5a +  6'  an<"  a2  — 4' 

12.    T,  — — T,  and  -j — T . 

a;  —  1    ar— 1  x3  — 1 

-n  a  &  m  —  n  a  +  b 

a  m  —  b  m  +  a  n  —  b  n'  2  a2  —  2  a  b  '         3  b  m  +  3  b  n 

14.  Reduce  ; r^—    -  ,  — ;— - ,  and 


(a-b)  (a-c)  '  (b  —  a)  (b-e)'  (c-a)(c-b) 

to  u  common  denominator. 


FRACTIONS.  77 

The  fractions  may  be  written  (Art   145)  as  follows : 

,  and 


(a  -  6)  (a  -  c)  '  (a  -b)(b-  e)  '  (a  -  c)  (6  -  c) 

The   least    common    denominator    is    now    (a  —  b)  (a  —  e) 

(b  —  c).     Applying  the  rule,  we  have  the  results, 

Q-a)(b-c)  (h-l)(a-c)       ^    ^ 


(a  -b)  (a-  c)  (b  —  c)'  (a-  b)  (a  -  c)  (b  -  c) 
(l-c)(a-b) 


(a  -b){a-  c)  (b  -  c) 
Kecluce  to  a  common  denominator  : 
,_        3  2  .a  —  2 

3 


,  Ans. 


a 

— 

V 

a  + 

V 

1 

2- 

X 

1 

+ 

X 

c 

> 
x  — 

+  d 

1' 

16.    z ,  7,  and 


x' 


1  —  x  .  b  —  a 

17.    -. j^-. pr-,  -p. sT? ~ ,  and 


(a  +  b)(a-b)'   (b-a)(c-d)'  (d-c)(a  +  b) 

153.  A  fraction  may  he  reduced  to  an  equivalent  one  hav- 
ing a  given  denominator,  by  dividing  the  given  denominator 
by  the  denominator  of  the  fraction,  and  multiplying  both  terms 
by  the  quotient. 

1.    Eeduce    — ry    to  an  equivalent  fraction  having 

ar  —  a  b  +  b" 

a3  +  b3  for  its  denominator. 

(a*  +  b3)  -j-  (a2  -ab  +  b2)  =  a  +  b; 
multiplying  both  terms  by  a  +  b, 

a  —  b  (a  -  b)  (a  +  b)  a2-b2 

a*-ab  +  b*  ~  (a2  -ab  +  b2)  (a  +  b)  ~  a3  +  b3,      **' 


78  algesra. 

EXAMPLES. 

2.  Reduce to  a  fraction  with  a2  —  b2  for  its  denom- 

a  +  b 

inator. 

x  -\-  1 

3.  Reduce   -   to  a  fraction  with  x2  +  5  x  —  24  for  its 

x  —  3 

denominator. 

.     -r,   ,  3  m  +  2 

4.  Reduce to  a  fraction  with  G  m?  —  19  m  +  10 

for  its  denominator. 

4 

5.  Reduce to  a  fraction  with  a3  —  b3  for  its  denom- 

a  —  0 

inator. 

6.  Reduce  1  +  x  to  a  fraction  with  1  —  x  for  its  denomi- 
nator. 


ADDITION  AND  SUBTRACTION   OF  FRACTIONS. 

154.     1.    Let  it  he  required  to  add  -  to  -. 

c         c 

In  --  and  -  ,  the  unit  is  divided  into  c  equal  parts,  and  a 

and  b  parts,  respectively,  are  taken,  or  in  all  a  +  b  parts ;  that 

a  +  6 

is    .     Ihus, 

c 

a      b        a  +  b 

-  +  -  =— -I— . 
c      c  c 

2.    Let  it  he  required  to  subtract  -  from  - . 

c  c 

The  result  must  he  such  a  quantity  as  when  added  to  7  will 

produce  -;    that  quantity  is  evidently  ■  •   (Art.    154,  1). 

,„,  a      b        a  —  b 

1  hus,  = . 

c       c  c 


Hence  the  following 


FRACTIONS.  79 

RULE. 

To  add  or  subtract  fractions,  reduce  them,  if  necessary,  to  a, 
common  denominator.  Add  or  subtract  the  numerators,  and 
write  the  result  over  the  common  denominator. 

The  final  result  should  be  reduced  to  its  simplest  form, 
wherever  such  reduction  is  possible. 

3b -a      b  +  a  .    1  -  4  o2 

1.    Add  — - ,       n  7     ,  and  — - — — . 

3a     '      2b  iab 

The  least  common  multiple  of  the  denominators  is  12  a  b. 
Then,  by  the  rule  of  Art.  152, 

3b  -a       b  +  a       1  -  4  b*        12  b2  -  4  ab       6  a  b  +  6  a2 

+  "ITT"  +        ,        ,        =  "    -To-T-  -  + 


+ 


3a  2  b  ±ab  12  a  b  12  a  b 

3  -  12  lr        12  fr2-4  a  6  +  6  a  fr  +  6  a2  +  3-12  b2 
12  a  b  12  a  b 

6a2+2ab  +  S 


12  a  b 


,  Ans. 


n       n     i  4  flj  —  1     .  6  «  —  2 

«.    subtract   — ^ from    — . 

2  x  6  a 

The  least  common  denominator  is  6  ax. 

6a  —2      4  a;  —  1      12  a  a;  —  4  a:      12  ax  — S  a 


Then, 


3a              2a?  Gaa;                   6  a  x 

12  a .r  —  4 x  —  (12  ax  —  3  a)  12  a  a;  —  4  a;  —  12  a  x  +  3  a 

6  a  x  6  ax 

3  a  —  4  a; 


6  a  a; 


,  Ans. 


Note.  When  a  fraction  whose  numerator  is  not  a  monomial  is  preceded 
by  a  -  sign,  it  will  be  found  convenient  to  enclose  its  numerator  in  a  pa- 
renthesis before  combining  with  the  other  numerators.  If  this  is  not  done, 
care  must  be  taken  to  change  the  sign  of  each  term  in  the  numerator  before 
combining. 


80  ALGEBRA. 

4«2-l       3«i2-2      5«2c2+3 


3.    Simplify 


2  a  c  3  b2  c  5  a 


3 


The  least  common  denominator  is  30  a  b2  c3. 

4a2-l    _  3  a  b2 -  2  _  5a2c2+3 
2  a  c  3  b2  c  5  a  c3 

60  a2  b2  c2  - 15  b'2 c2 _  30  a2  b2  c2 -  20  a  c2      30  a2  b2  c2  +  18  If 
30  a  b2  cs  30  d  b2  c3  30  a  6a  c3 

_  60  a2  b2  c2  - 15  b2  c2  -  (30  a2  b2  c2-20a  c2)  -  (30  a2  b2  c2  +18  b?) 

30  a  62  c3 

_  60  a2  &2  c2 -  15  62  c2 -  30  a2  b2  c2  +  20  a  c2 -  30 a2b2 c2  - 18  b2 

30  a  6'2  c3 

20  a  c2  - 15  62  c2  - 18  b2 


30  a  b2c3  >AnS' 


EXAMPLES. 

Simplify  the  following : 

.     2x  —  5      3  a  + 11  _    a  —  b      2a  +  b      b  —  3a 

4. \- .  9. 1 1 . 

12              18  4^6               8 

3              1  a2  + 1      6  a3  + 1       6-2 

*     5  a  J2  +2a2i'  '3  a2           12  a3     +  ITT' 

2  a  +  3      3  a  +  5  2a;-l      2x  +  3      6cc  +  l 

6                8      '  "12           ~H~       ^2(P" 

ra  —  2      2  — 3?»,?i2  m  +  2      m  +  2      m  +  3 

'    2m »         3  m2 n3     '  '     ~J~        ~U~      ~2lT' 

b  —  4a      a  +  5b  10    2      2x  —  1      3.r2+l 

o.    — — 1 — — - — .  lo. 


24:  a    ^    30b    '  '3         6x  9 a;2     ' 

,.     a  — 2      3cc  +  l      6  a: -5      3 

14.   ■  -\ ' 

2      +      3  4  5 

3»  +  l      2&-1      4<?-l      6^+1 
12  a        ~~8lT"  +    16  c  24^' 


16.    Simplify 


FRACTIONS.  81 

2x  +  l  3x  — 1  11 


2  x  (x  —  1)      3  a;  (a?  +  1)      4  (a-2  -  1) 
The  least  common  denominator  is  12  x  (x2  —  1). 
2x  +  l  3x  —  1  11 


Then, 


2  x  (x  -  1)      3  x  (x  +  1)      4  (a;2  -  1) 
6  (a  +  1)  (2  x  +  1)      4  (as  - 1)  (3  x  —  1)  33  a; 


12x(x2-l)  12x(x2-l)  12x(x2-l) 

12  x2  +  18  x  +  6      12  x-  -  16  x  +  4  33  x 


12  x  (x2  -  1)  12  x  (x-  -  1)         12  x  (x2  -  1) 

_  12  x2  +  18  x  +  6  -  (12  x3  -  16  x  +  4)  -  33  x 
12  x  (x2  -  1) 

x  +  2 


12  x  (x2  -  1) 


Aiis. 


Simplify  the  following : 

»;_*   +   *_.  »i±»+s=» 

x  +  2      3  —  x  a  —  6       a  +  6 

18.  -i L_.  20.^-^. 

x  +  7      x  +  8  1  —  xl  +  x 

a  />  2  a  5 

«1.    — —7  H j  H — o To  • 

a  +  0      a  —  0      a'  —  b" 

1  1  2x 

22.   + 


x  +  y       x  —  //       x'  +  y- 

1  x  3 

23.   5 0-^r  + 


24. 


X  —  1        x2— 1       X3  —  1 

2  x  —  6  x  +  2  x  + 1 


x2+3x  +  2      x2-2x  — 3      x1—  x—  6' 


x  x  2  x 


25.    Simplify  — — r  +  — h  -5 


x  +  1       1  —  x       x'2  —  1 


82  ALGEBRA. 

The  expression  may  be  written  (Art.  143)  as  follows : 

X  X  Li  X 

+ 


X  +  1        x  —  1        X*  —  1 
The  least  common  denominator  is  ar  —  1. 

ihen>    ZTT^T-Z. T  +  ~2 7  =  31 i—  Z72 T  +  Z2- 


x  +  1       x  —  1       ar  —  1      a;'2  —  1      x'  —1      x2  —  1 
or  —  a:  —  (.t2  +  cc)  +  2  a;  0 


x1  —  1 


Simplify  the  following : 


x* 


=  0,  Ans.  (Art.  102). 


3  4 

26.    -2-+*.  28.   -^-_+  — 


a  —  &      b  —  a  3  x  —  a2      #2  —  9 

_,_.     o  rt  -p  x      o  a  —  J.  -.-.         x  x  x 

27. 1 — .  29. 1 

3a+3^2-2a  1+x       1-x 

1  1  1 

30.   T-^f-Ti z  + 


x 


31. 


(a  —  b){b  —  c)      (b  —  a)  (a  —  c)      (c  -  a)  (c  —  b) 
2  3  1 


(a;  -  2)  (a;  -  3)      (3  -*)  (4  -a;)      (a;  -  4)  (2  -a;)  ' 


MULTIPLICATION   OF   FRACTIONS. 

155.  We  showed,  in  Art.  137,  that  a  fraction  could  be 
multiplied  by  an  integer  either  by  multiplying  its  numerator 
or  by  dividing  its  denominator  by  that  integer.  We  will  now 
show  how  to  multiply  one  fraction  by  another. 

Let  it  be  required  to  multiply  -  by  - . 

Let  -  =  x,  and  -  =  y ; 

where  x  and  y  may  be  either  integral  or  fractional. 


FRACTIONS.  83 

Since  the  dividend  equals  the  product  of  the  divisor  and 

quotient. 

a  =  b  x,  and  c  =  d  y. 

Therefore,  hy  Art.  44,  Ax.  3,  a  c  =  b  d  x  y. 

Regarding  a  c  as  the  dividend,  b  d  as  the  divisor,  and  x  y  as 

the  quotient,  we  have 

a  c 

xy  =  Vd- 

Therefore,  putting  for  x  and  y  their  values, 

a      c      a  c 
~bXd^bd' 
Hence  the  following 

RULE. 

Multiply  the  numerators  together  for  the  numerator  of  the 
resulting  fraction,  and  the  denominators  for  its  denominator. 

Mixed  quantities  should  be  reduced  to  a  fractional  form 
before  applying  the  rule. 

When  there  are  common  factors  in  the  numerators  and 
denominators,  they  should  be  cancelled  before  performing  the 
multiplication. 

EXAMPLES. 

-     ,,  1A.  -■  ,       6x2y     10  a2  y         .  3b*x* 

1.    Multiply  together  5 — ^  ,  -=-= — -  ,  and  -. ~  . 

1  J      &  5  a3  b2 '     9  b  x    '  4  a  y2 

6x2y      10  a2y      3  J4  xs     6  x  10  X  3  a2  b*  x5  y2     b  x*      . 
5a3b2        9bx    A  lay2    '  5  X  9  X  4  «4  b3  x  y2  '  '  a2  '  " 

Multiply  together  the  following  : 

a2  5  c       -.    a3b2  .     3  abx2  5  x  y2 


•   — 2  and  —3 — j  •  4-  -5 — 2-  ancl  5 — r- 

ra  ?r          md  n  d  bay2,  0  ab  x 

0     3  a3x       ,4fflJ  _  m  ?/n        ,    a  x 

3'   ^li-  and  KTZ,  ■  5-  7^1  and 


7  A4  5Am  4 « x  m  yn 


84  ALGEBEA. 

e     2a     6c        ,5b                   0    3ab2    3ac2  ,Sa<P 

o6      5a           be                          4cd     2&d  9&c 

„    8  a--     15  y2         ,3  s;4             _     3  m*     2  n*  .  11  z2 

'•    fT~3>    i7 — 5  ?  and  in    q  •  "•      ,,     o  >    o — .and-; — 5-. 

9  if    lb  zs'         10  cc3                 2a;2'3??i'  4w2 

10.    Multiply  together 

ar2_2cc         x2-9  .        a2  +  a; 

,  and 


a;2_2a;-3'    a;2-*'  cc2+a:-.6' 

r-2x         x2-9  cc2  +  ^ 

X-s X 


ce2  —  2  cc  —  3      x-  —  x      x2  +  x  —  6 

x(x-2)(x  +  3)  (a  -  3)  x  (x  +  1)  x 

(x-3)(x  +  l)x(x- 1)  (a  +  3)  (x -2) ~ x~^l ' 

Multiply  together  the  following : 

1t     3.r2-cc        ,  10 

11.    = and 


5  2cc2-4a;' 

,.    4  a; +  2  5cc 

1*.    — ^ and 


2  x  +  1 ' 

1Q     a2-2ab  +  b2       .  b 

lo. ; and 


a + b  ax  —  bx 

., .       a  —  b  .     a2  —  b2 

14.   -=—   -—  and 


a2  +  a  b  a2  —  a  b 

,  _     1  —  x2    1  —  y2         , 
10.    q ,  s ,  and 


1  +  1/  '  X  +  X2'  1  —  X 

.0     x2-W        .  x2-25 
lb.      „        —  and  -5 - —  . 

Xs  +  ox  r-4a; 

a3  —  a2  +  a  xs  —  8 

17.    -r and  — 5  • 

x2  +  2  x  +  4  as  +  1 

,_     a:2  +  5.r+G        ,  x2—  Ix 
18>   x^I^^Yl  and  ^4' 


FRACTIONS.  85 


4       5  x 

19.    1  H —  and  — 


x      x1  x2  —  8x  +  7 


20.   -4-1  and 


«X/  fcC 


2-5cc  +  6" 


a,a  _  3  x  +  2      a2  -  7  a;  +  12  a3  -  5  x2 

2L   aj»_8aj  +  15'    *2-5x  +  4'  and    a* -4  ' 


22.   ^ 2— j  Lj^r>  and  1  +  ^— 

a;  —  x  y  +  y     x~  +  x  y  +  y  x  —  y 

a*  _  U1  -  c2  +  2  5  c  a2-p-c2-2bc 

i6-    a-  +  c--b2  +  2ac  a2  +  c2-b2-2ac' 

OA     a  +  b      a  —  b        4  b2  a  +  b 

a  —  b      a  +  b      a  — b  Jo 

_„    2x-\-y      .  y  x2  ..  x2  —  y2 

25.   —  —  —  1 ^ = r-  and  -= — V- 

x  +  y  y  —  cc       x*  —  y*  x'  +  y 


DIVISION   OF  FRACTIONS. 

156.  We  showed,  in  Art.  138,  that  a  fraction  could  be 
divided  by  an  integer  either  by  dividing  its  numerator  or  by 
multiplying  its  denominator  by  that  integer.  We  will  now 
show  how  to  divide  one  fraction  by  another. 

Let  it  be  required  to  divide  -  by  - . 

ft  n 

Let  x  denote  the  quotient  of  --$-—. 

b      a 

Then,  since  the  quotient  multiplied  by  the  divisor  gives  the 
dividend,  we  have 

c      a  xe     a 

XXd  =  b]  °r'  ~d=b' 


86  ALGEBRA. 


Multiplying  each  of  these  equals  by  -  (Art.  44,  Ax.  3), 

G 


Therefore, 


a  d 
x  = 

be 

a 

c      ad 
d       be' 

Hence  the  following 


RULE. 

Invert  the  divisor,  and  proceed  as  in  multiplication. 
Mixed   quantities   should  he   reduced   to  a  fractional  form, 


before  applying  the  rule. 


EXAMPLES. 

6  aH    ,         9ab3 


1.   Divide   r    .    ,  by  -T7r — = — r. 
5  xs  y*     J    10  x2  y5 


6a2b        9ab5         6a2b      10  x2  yh      Any      . 

: -^ y  l_  — "L      Ant 

5a:3?/4*  10.x2?/5      5xsy**   9abs        3b2  x' 

x2  —  9        x  +  3 
2.   Divide  —zrs —  by  — p — . 

x2  —  9     x  +  S   _(x  +  3)(x- 3)  5      _x  —  S 

~TE~  ''    ~5~~~~      ~16~       ~Xx~+3~~ 3~ ' 

Divide  the  following: 

7  ??i2         3n2  x2  —  y2  x2+xy 

6-   ~2~    y  "13" '  ^-2^  +  y2    y  ar-y  * 

.       7«3i     .      14  a  b*  _    n         Sy2  0         5  V 

1 J  wr  ?r  o  ra  w  xz  —  y  x  —  y 


5. 

18  7??-  a:8  1      6  m2  a:4 
25  wy2      }    5«22/6' 

-1.1 

*■  a2+2a-15    }   a2-2a-3' 

6. 

1       4t          x2      x 
4"^by  12+3- 

in      a;3  —  4  a*     ,     x2—3x-\-2 
'  a;2  +  5x  +  6  b}  «a+2aj-3' 

FRACTIONS.  87 


COMPLEX   FRACTIONS. 

157.  A  Complex  Fraction  is  one  having  a  fraction  in  its 
numerator,  or  denominator,  or  both.  It  may  be  regarded  as  a 
case  in  division,  since  its  numerator  answers  to  the  dividend, 
and  its  denominator  to  tbe  divisor. 

However,  since  multiplying  a  fraction  by  any  multiple  of 
its  denominator  must  cancel  that  denominator,  to  simplify  a 
complex  fraction,  we  may  multiphj  both  of  its  terms  by  the 
least  common  multiple  of  their  denominators. 

EXAMPLES. 

a 
1.    Reduce   -  to  its  simplest  form. 


FIRST   METHOD. 

Proceeding  as  in  division, 

a 

c      a      b       a  b 
-=^Xj=—t,  Ans. 
a      c      a      ca 

b 

SECOND    METHOD. 

Multiplying  both  terms  by  the  least  common  multiple  of 
their  denominators, 

a      a 

-Xbc 
e       c  a  o       . 

-=- =  —T)  Ans. 

da,         c  a 

»   bxbe 

a  a 


2.    Reduce  — ^—  —  to  its  simplest  form. 

b  a 

a  —  b      a  +  b 


88  ALGEBEA. 

The  least  common  multiple  of  the  denominators  is  a2  —  b2. 
Multiplying  each  term  by  a'2  —  lr,  we  have 

a  (a  -\-b)  —  a  (a  —  b)      cr  +  ab  —  a2  +  ab        2  a  b 

Ans. 


b  (a  +  b)  +  a  (a  —  b)      a  b  +  b~  +  a2  —  a  b      a1  +  b2 ' 
3.    Reduce - —  to  its  simplest  form. 

x 

1  1  X+  1  33  +  1 

,  Ans. 


1  1  _,  a*  a:  +  1  +  a;      2cc  +  1 

.1  a;  +  1 

1  +- 

Reduce  the  following  to  their  simplest  forms : 
4.   -L-.  8.   !-|.  12 


5. 

b 

a-\ — 
c 

m 

x 

n 

6. 

n 
»-g 

cc 


y-^  +  9 

7.       -5=-4.  11.  15 


1  * 

1  + 

X 

1 

x2  + 

X 

1  ' 

1  + 

X 

a 

b 

b 

a 

1 

1* 

b 

a 

1 

x2  + 

2 

X- 

— 

2/ 

w  +  ?i  1  «-  +  ft  b  +  .b2 ' 


n        12 

x  —  7  -\ 

9.  j.  13.    -, 

a_q      18 

a;  +  3 

X 


10. 


14. 

1 

1  —  X 

1 

1  +  £C 

1 

=-    -H 

1 

-= 

31 

1+  -c  +  1 
3  —  x 


SIMPLE  EQUATIONS.  89 

a2  +  b2      a2  —  lr  in  —  n      m3  —  ns 

,n     a2  —  b2      a2  +  b2  10     m  +  n      in3  +  n3 

16.    ; =—  •  !"•    ~? '?• 

a  +  b      a  —  b  m  +  n      rnr  +  nr 


+ 
a  —  b      a  +  b  m  —  n      mr  —  n 


.2 


i7.     x  +  y    y  ,         i9. 


•>   «2 


x  +  2  y         x  2 " 

a; 


v/  a;  +  y  a  +  x 

158.     In  Art.  42,  we  defined  the  reciprocal  of  a  quantity 

as  being  1  divided *by  that  quantity.     Therefore  the  reciprocal 

in  it 

of  —  =  —  =  —  ;  or,  the  reciprocal  of  a  fraction  is  the  frac- 
n       in       m 

ii 
tion  inverted. 


XII.  — SIMPLE  EQUATIONS. 

159.  An  Equation  is  an  expression  of  equality  between 
two  quantities.     Thus, 

x  +  4  =  16 

is  an  equation,  expressing  the  equality  of  the  quantities  x  +  4 
and  16. 

160.  The  First  Member  of  an  equation  is  the  quantity  on 
the  left  of  the  sign  of  equality.  The  Second  Member  is  the 
quantity  on  the  right  of  that  sign.  Thus,  in  the  equation 
x  +  4  =  16,  x  +  4  is  the  first  member,  and  16  is  the  second 
member. 

The  sides  of  an  equation  are  its  two  members. 

161.  An  Identical  Equation  is  one  in  which  the  two  mem- 
bers are  equal,  whatever  values  are  given  to  the  letters  in- 
volved, if  the  same  value  be  given  to'  the  same  letter  in  every 
part  of  the  equation  ;  as, 


90  ALGEBRA. 

2a  +  2bc  =  2(d  +  bc). 

162.  Equations  usually  consist  of  known  and  unknown 
quantities.  Unknown  quantities  are  generally  represented  by 
the  last  letters  of  the  alphabet,  x,  y,  z;  but  any  letter  may 
stand  for  an  unknown  quantity.  Known  quantities  are  repre- 
sented by  numbers,  or  by  any  except  the  last  letters  of  the 
alphabet. 

163.  A  Numerical  Equation  is  one  in  which  all  the  known 
quantities  are  represented  by  numbers  ;  as, 

2  x  —  11  =  x  —  5. 

A  Literal  Equation  is  one  in  which  some  or  all  the  known 
quantities  are  expressed  by  letters  j  as, 

2x  +  a  =  bx2  —  10. 

164.  The  Degree  of  an  equation  containing  but  one  un- 
known quantity  is  denoted  by  the  highest  power  of  that 
unknown  quantitj'  in  the  equation.     Thus, 

>  are  equations  of  the  first  degree. 
and  c  x  =  a'2  +  b  d '  ) 

3  x2  —  2  x  =  65  is  an  equation  of  the  second  degree. 

In  like  manner  we  have  equations  of  the  third  degree,  fourth 
degree,  and  so  on. 

When  an  equation  contains  more  than  one  unknown  quan- 
tity, its  degree  is  determined  by  the  greatest  sum  of  the 
exponents  of  the  unknown  quantities  in  any  term.     Thus, 

x  +  x  y  =  25  is  an  equation  of  the  second  degree. 

a;2  —  y2  z  =  a  b3  is  an  equation  of  the  third  degree. 

Note.  These  definitions  of  degree  require  that  the  equation  shall  not 
contain  unknown  quantities  in  the  denominators  of  fractions,  or  under 
radical  signs,  or  affected  with  fractional  or  negative  exponents. 


SIMPLE   EQUATIONS.  91 

165.  A  Simple  Equation  is  an  equation  of  the  first  degree. 

166.  The  Root  of  an  equation  containing'  but  one  unknown 
quantity  is  the  value  of  that  unknown  quantity;  or  it  is  tin- 
value  which,  being  put  in  place  of  the  unknown  quantity, 
makes  the  equation  identical.     Thus,  in  the  equation 

3x  —  7  =  x  +  9, 
if  8  is  put  in  place  of  x,  the  equation  becomes 

24  -  7  =  8'+  9, 
which  is  identical;  hence  the  root  of  the  equation  is  8. 

Note.     An  equation  may  have  more  than  one  root.     For  example,  in 

the  equation 

x2  =  7  J! -12, 

if  3  is  put  in  place  of  x,  the  equation  becomes  9  =  21-12;  and  if  4  is  put 
in  place  of  x,  it  becomes  16  =  28  —  12.  Each  of  these  results  being  iden- 
tical, it  follows  that  either  3  or  4  is  a  root  of  the  equation. 

167  It  will  be  shown  hereafter  that  a  simple  equation  has 
but  one  root;  an  equation  of  the  second  degree,  two  mots; 
and,  in  general,  that  the  degree  of  the  equation  and  the  num- 
ber of  its  roots  correspond. 

168.  The  solution  of  an  equation  is  the  process  of  finding 
its  roots.  A  root  is  verified,  or  the  equation  satisfied,  when, 
the  root  being  substituted  for  its  symbol,  the  equation  becomes 
identical. 

TRANSFORMATION   OF  EQUATIONS. 

169.  To  Transform  an  equation  is  to  change  its  form  with- 
out destroying  the  equality. 

170.  The  operations  required  in  the  transformation  are 
based  upon  the  general  principle  deduced  directly  from  the 
axioms  (Art.  44)  : 


92  ALGEBRA. 

If  the  same  operations  are  performed  upon  equal  quantities, 
the  results  will  be  equal. 

Hence, 

Both  members  of  an  equation  may  be  increased,  diminished, 
multiplied,  or  divided  by  the  same  quantity,  without  destroy- 
ing the  equality. 

TRANSPOSITION. 

171.  To  Transpose  a  term  of  an  equation  is  to  change 
it  from  one  member  to  the  other  without  destroying  the 
equality. 

172.  Consider  the  equation  x  —  a  =  &. 
Adding  a  to  each  member  (Art.  170),  we  have 

x  —  a  +  a  =  b  +  a 
or,  x  =  b  +  a, 

where   —  a  has  been  transposed  to  the  second   member   by 
changing  its  sign. 

173.  Again,  consider  the  equation  x  +  a  =  b. 
Subtracting  a  from  each  member  (Art.  170),  we  have 

x  +  a  —  a  =  b  —  a 

or,  x  —  b  —  a, 

where  a  has  been  transposed  to  the  second  member  by  chang- 
ing its  sign. 

174.  Hence  the  following 

EULE. 

Any  term  may  he  indisposed  from  one  member  of  an  equa- 
tion  t<>  the  other,  provided  its  sign  be  changed. 

Also,  if  the  same  term  appear  in  both  members  of  an  equa- 
tion affected  with  tin  same  sign,  it  may  be  suppressed. 


SIMPLE   EQUATIONS.  93 

1.    In  the  equation    2x  —  12  4-  '3  =  a;  —  5a  +  9,    transpose 

the  unknown  terms  to  the  first  member,  and  the  known  terms 
to  the  second. 

Eesult,  2x  —  x  +  5  a:  =  12  —  3  +  9. 

EXAMPLES. 

Transpose  the  unknown  terms  to  the  first  member,  and  the 
known  terms  to  the  second,  in  the  following : 

2.  3x  —  2a  =  45+.2x. 

3.  4:X  +  9  =  25-12x. 

4.  4 a2 x  +  b2  =  —  4  a  bx+±ac+  b2. 

5.  a  c  +  c  x  —  a  d  =  2  a  —  7  #. 

6.  &  c  +  a2  x  —  m  ?r  =  b  x  +  a  tZ  —  5. 

7.  3  —  &  —  x  =  c  —  3x. 

8.  2a  —  3c  =  5ic  —  b  —  dx. 

9.  10  jc  -  312  =  32  x  +  21  -  52  x. 

CLEARING  OF  FRACTIONS. 

175.     1.   Clear  the  equation  -jr T  =  -X-+  s  of  fractions. 

o        4        o        o 

The  least  common  multiple  of  3,  4,  G,  and  8  is  24.     Multi- 
plying each  term  of  the  equation  by  24  (Art.  170),  we  have 

16  x  -  30  =  20  x  +  9, 

where  the  denominators  have  been  removed.     Hence  the  fol- 
lowing 

RULE. 

Multiply  each  term  of  the  equation  by  the  least  common 
multiple  of  the  denominators. 


2, 

ax             (I  x       m 

c  — 

b                 en 

x        2  a          1          x 

:;. 

2  a      3  b       lab      6 

ax       ex       a 

4. 

x      i    +    ,        —  o 

b           a        e 

94  ALGEBRA. 

Note.  The  operation  of  clearing  of  fractions  may  be  performed  by- 
multiplying  each  term  of  the  equation  by  any  common  multiple  of  the  de- 
nominators. The  product  of  all  the  denominators  is  obviously  a  common 
multiple,  and  the  rule  is  sometimes  given  as  follows  :  "Multiply  each  term 
of  the  equation  by  the  product  of  all  the  denominators." 

EXAMPLES. 

Clear  the  following  equations  of  fractions: 

3x  5x  b 

4  0  3 

7.  x-%+ 20  =  1  +  ^  +  26. 

n      w  X         O  0         ox  n 

8'  is-^-ir+!"l=()- 

—,  *As  its  %K/  C\C\  f\  *-^   *^  *^  HO  J-O   </• 

b'  5  +  L2  =  l0_"-  "    12"       3T~lo  =  8~~6~- 

5  x 5      -\^ q  x     9J  i_j  x 

10.    Clear  the  equation  21  —  ■ — ==  — — —  - -t 

o  1G  2 

of  fractions. 

The  least  common  denominator  is  16  ;  multiplying  each  term 
by  1G,  we  have 

336  -  (10  x  -  10)  =  11  -  3  x  -  (776  -  56  x) 
or,  : !: U ;  -  lO  x  +  10  =  11  -  3  x  -  776  +  56  x,  Am. 

Note.  When  a  fraction,  whose  numerator  is  not  a  monomial,  is  preceded 
by  a  —sign,  it  will  he  found  convenient,  on  clearing  of  fractions,  to  enclose 
the  numerator  in  a  parenthesis.  If  this  is  not  done,  care  must  be  taken 
to  change  the  sign  of  each  term  in  the  numerator. 

Clear  the  following  equations  of  fractions : 

11     x      a  +  x 15  12    ax  +  b      cx  +  d a 

2  3~~    ~~2'  r  be        '  V 


SIMPLE  EQUATIONS.  95 

13.   -1 -1^  =  0.  15.      ?_ ?___5a!_0 

1  +  a;      1  —  a?  x  +  1      a;  —  1      ar  —  1 

14    a       ar  —  3       1  _a  16  ,T  ~*~  ^      a?  —  3      2«  +  l_ 

2~2x  +  l_3_    '  *  ~5~      ~2~        ~3~ 


CHANGING   SIGNS. 

176.  The  signs  of  all  the  terms  of  an  equation  may  be 
changed  without  destroying  the  equality. 

For,    in   the    equation  a  —  x  =  b  —  c,  let   all  the  terms  he 
multiplied  by  —  1  (Art.  170).     Then, 

—  a  +  x  =  —  b  +  c 
or,  x  —  a  =  c  —  b. 

For  example,  the  equation  —  5  x  —  a  =  3  x  —  b,  by  chang- 
ing the  signs  of  all  the  terms,  may  he  written 

5x  +  a  =  b  —  3x. 

SOLUTION   OF   SIMPLE   EQUATIONS. 

177.  To  solve  a  simple  equation  containing  hut  one   un- 
known quantity. 

1.   Solve  the  equation  5  x  —  7  =  a?  +  9. 

Transposing  the  unknown  terms  to  the  first  member,  and 
the  known  terms  to  the  second, 

5  x  —  ^  =  7  +  9 
Uniting  similar  terms,  4  x  =  16 

Dividing  each  member  by  4  (Art.  170), 

x  =  4,  Ans. 

This  value  of  x  we  may  verify  (Art.  168).     Thus,  substi- 
tuting 4  for  x  in  the  given  equation,  it  becomes 

20  -  7  =  4  +  9, 

which  is  identical ;  hence  the  value  of  x  is  verified. 


96  ALGEBRA. 

2.    Solve  the  equation  8  x  +  19  =  25  x  —  32. 

Transposing,  8  x  —  25  x  =  —  19  —  32 

Uniting  terms,  —  17  x  =  —  51 

Dividing  by  —  17,  cc  =  3,  Ans. 

To  verify  the  result,  put  3  for  x  in  the  given  equation. 

Then,  24  +  19  =  75  -  32 

or,  43  =  43. 

o    a  i      .i  .-       3x       5      2x      x 

o.    bolve  the  equation  — — |-  -  =  — . 

4  0  o  *j 

Clearing  of  fractions,  by  multiplying  each  term  of  the  equa- 
tion by  12,  the  least  common  multiple  of  the  denominators, 

9x  +  10  =  8x  —  Gx 

Transposing,  9a;  —  8x  +  6x  =  —  10 

Uniting  terms,  7  x  =  —  10 

Dividing  by  7,  x  =  — — ,  Ans. 

To  verify  this  result,  put  x  = =r  in  the  given  equation. 

Then,  _30      5_    _20      10 

_.28  +  6~-21+14 


or, 


or, 


-90  +  70    _-80  +  60 
~84_  "SlT 

_20_    _20 

84 "       84' 


RULE. 

Clear  the  equation  of  fractions  if  it  has  any.  Transpose 
the  unknown  terms  to  tit  e  first  member,  and  the  known  terms 
to  the  second,  and  reduce  each  member  to  its  simples}  firm. 
Diride  both  members  of  the  resulting  equation  by  the  coefficient 
of  the  unknown  quantity. 


SIMPLE   EQUATIONS.  97 

EXAMPLES. 

Solve  the  following  equations  : 

4.  3sc  +  5  =  a;  +  ll.  7.  3x  +  2-5ce  =  :b-7  +  3. 

5.  3z-2:=5x-lG.  8.  18-5cc-2x  =  3  +  a;  +  7x. 

6.  2-2a;  =  3-a?.  9.  5x^3  +  17  =  19-2x-2. 

10.    Solve  the  equation 

5(7  +  3x)-(2a;-3)(l-2x)-(2a;-3)2-(5  +  :*0=O- 
Performing  the  operations  indicated,  we  have 

35  +  15a:  +  4a;2-8:r  +  3-4x2  +  12:r-9-5-a;  =  0 

Transposing,  and  suppressing  the  terms  4  x2  and  —  4  x2, 

15  ;c-8;c  + 12  a:-a;  =  - 35 -3  +  9  +  5 

18  x  =  -  24 

24  4     , 

x  =  -lS  =  -3>AnS- 

Solve  the  following  equations  : 

11.  3  +  2  (2x  +  S)  =  2x  -3(2  x  +  1). 

12.  2cc  —  (4a;-l)=5a;-0-l). 

13.  7  («-2)  -5  (a  +  3)  =  3  (2x-  5)  -6  (4a;-  1). 

14.  3(3x  +  5)-2(5z-3)=13-(5;c-16). 

15.  (2  x  - 1)  (3  a;  +  2)  =  (3  a;  -  5)  (2  a  +  20). 

16.  (5  -  G  a-)  (2  x  - 1)  =  (3  x  +  3)  (13  -  4  x). 

17.  (^-3)'2-(5-a-)2  =  -4'a;. 

18.  (2a;-l)2-3(^-2)  +  5(3x-2)-(5-2a;j2  =  0. 

•       3        7         7         5 

19.    Solve  the  equation „ —  =  zr^  —  k~ -• 


98  ALGEBRA. 

Clearing  of  fractions,  by  multiplying  each  term  by  12  x,  the 
least  common  multiple  of  the  denominators, 

36  -  42  =  7  x  -  20 

-  7  x  =  -  36  +  42  -  20 

-  7  a;  =  -  14 

Solve  the  following  equations : 

20.   i^_7=— -—        24    ^—x-2x      8x      11 
4  3  4    '  *     5       *-^--2— -11. 

2i     1,     1        1         1  ok    x      -^      x  _x      3x 

6+2^_4+12^-  Mt   2  +  ~6~3  =  6~7T- 

«.  |-|+|  =  18.  26.*-f  +  20  =  |+|  +  26. 

23.  |_?_*=I_i  27.  2-^  =  7-   3 


345a?  a;      2a;  2a;' 

oqqt      n  .  •       3a;  —  1      2  a;  +  1      4a;  —  5 

ao.    holve  the  equation  : =  4. 

4  3  5 

Multiplying  each  term  by  60, 

45  x  — 15  -  (40  x  +  20)  -  (48  x  -  60)  =  240 

45  x  —  15  -  40  x  —  20  —  48  x  +  60  =  240 

45  x  -  40  x  —  48  x  =  15  +  20  -  60  +  240 

-  43  a;  =  215 

X  =  —  5,    ^1?2S. 

Solve  the  following  equations  : 

oq    q         5  ./■  +  3      7x  OA  2  .r  + 1      r         5 

29.   3>x-\ —  =  -s-.  30.   x =—  =  5x  —  -=. 

7-2  5  3 

31.    7  a- 7 =  3  x  +  7. 


SIMPLE  EQUATIONS.  99 


32.    2-7^~^  =  3x 


33. 
34. 


6  4 

5a--2_3a-  +  4      7x-\-2_x  — 10 
"IT  ~T~  ~6~  ~2~ 

a;  +  1      2  a;  —  5  _  11  x  +  5      a;  —  13 
~2~         "5"  ~W         ~3       ' 


5a-  +  l      17  x  +  7      3.T-1      7  a;  — 1 
4  +  a-      3  a;  —  2      11  a-  +  2      2  -  9  a; 


36. 


14 


2  a-  +  1      4 a-  +  5      8  +  x      2x  +  5 
'     ~3~~  ~T~        ~~6_        ~8~ 

2  3  1 


38.    Solve  the  equation 


x  —  1      a1  +  1      a'~  —  1 " 

Clearing  of  fractions,  by  multiplying  each  term  by  x2  —  1, 
2  (a-  +  1)  -  3  (x  -  1)  =  1 
2a;  +  2-3a-  +  3  =  l 

2a-3a;  =  -2-3  +  l 
—  a:  =  —  4 
x  =  4,  ^4«s. 
4  a-  +  3      12  a-  -  5      2  x  -  1 


39.    Solve  the  equation 


10  5  a;  — 1  "         5 


Clearing   of    fractions   partially,    by   multiplying   each    term 

by  10, 

0      120  a; -50      ,         0 

4a;  +  3 =-   — t—  =  ±x  —  2 

o .''  —  1 

_      ,         .     120  a; -50 

4a;  +  3  —  4.r  +  2  =  — = -.— 

ox  —  1 

:      120  x  -  50 

5  =  —= — 

o  a-  —  1 


100  ALGEBRA. 

Clearing  of  fractions,  by  multiplying  each  term  by  5  x  —  1, 

25  x  -  5  =  120  x  -  50 
25  a;  -  120  x  =  5  -  50 
-  95  £  =  -  45 
_45_    9 

x-y5-vJ>Ans' 

Note.  If  the  denominators  are  partly  monomial,  and  partly  polynomial, 
clear  of  fractions  at  first  partially,  multiplying  by  such  a  quantity  as  will 
remove  the  monomial  denominators. 

Solve  the  following  equations : 

1  —X      1  +  x      1  —  ar 
x  —  1      x  +  1  3 


X       X2  — 

5x     2 

3      j3  x  - 

-7      3' 

2x-l 

2*  +  7 

3a;  +  4~ 

"3^  +  2 

5-2a 

3-2x 

41  —  '  4fj 

*0'   x._2      s  +  2~x2-4' 

'   ~x~+Y''  :^+T*  9=    "3"     "  1-dx  ' 

„_   <6x2-3x  +  2      _  ._    2ar  +  3a;        1 

43-   rr^ s s  =  3.  47.   -^     — —  +  — -  =  x  +  1. 

2  a;2  +  5  a;  —  7  2  *  +  1        3  <c 

48.  Solve  the  equation  2  a  ce  —  3b  =  x  +  c  —  3  ax. 
Transposing  and  uniting  terms,  5  a  x  —  x  =  3b  -\-  c 
Factoring  the  first  member,       x  (5  a  —  1)  =  3  b  +  c 
Dividing  by  5  a  —  1,  a;  =  = r  ,  ^4»s. 

49.  Solve  the  equation  (&  —  c  .r)2  —  (a  —  c  x)'2  =  b  (b  —  a). 
Performing  the  operations  indicated, 

b2 -2bcx+  c2 x2 -  a2  +  2acx-  c2 x2  =  b2 -  a  b 

Suppressing  the   term    lr  in    both    members,  and    the    terms 
c2  x2  and  —  c2  x2  in  the  first  member, 


SIMPLE  EQUATIONS.  101 

—  2bcx  —  a2  +  2acx  =  —  ab 

2  a  ex  —  2b  cx  =  a2  —  ab 

Factoring  both  members,  2  c  x  (a  —  b)  =  a  (a  —  b) 

n      .         ,.  a  (a  —  b)        a       . 

Dividing  by  2e(a-b),       x=  2c^_^  =  ^  ,  Ans. 

Solve,  the  following  equations  : 

50.  2  ax  +  d  —  3c  —  bx. 

51.  2  x  —  Aa  =  3  ax  +  a2  —  a2x. 

52.  2  a  x  +  6  b2  =  3  b  x  +  4  a  b. 

53.  6  b  m  x  —  5  a  n  —  lo  a  ni  —  2bnx. 

54.  (or  -  2  x)2  =  (4  x  -  b)  (x  +  4  e). 

55.  (2  a  -  3  x)  (2  a  +  3x)  =  b2-(3x-  b)2. 

56.  (3  a  —  x)  (a  +  2  x)  =  (5  a  +  x)  (a  —  2  x), 

3b  x      2  _  3  _  2bx 

c       a         c 

-3+4^=3V2«(2-3«>- 


57. 

a 

X 

58. 

2a 

59. 

X 

2 

60. 

X 

a  b 

61. 

X 

2 

1  +  2  ax       2x  +  \ 


2  a  a2 

X  +  "  b  X 

~~3lT    =3T~ 


(«-!)• 


I>  c  x       x       a  c  —  4  6  a; 


2b  c  6  c  3b  c 

62.   Solve  the  equation  .2  x  -  .01  -  .03  x  =  .113  x  +  .161. 

FIRST    METHOD. 

('banging  the  decimals  into  common  fractions. 
2x        1        3x      118  a;       161 


10       100      100    "  1000       1000 


102  ALGEBRA. 

Multiplying  each  term  by  1000, 

200  x  -  10  -  30  x  =  113  x  +  161 
57  x  =  171 
x  =  3,  Ans. 

SECOND   METHOD. 

Transposing,  .2  x  —  .03  x  —  .113  x  =  .01  +  .101 
Uniting  terms,  .057  x  =  .171 

Dividing  by  .057,  x  =  3,  ^4ms. 

Solve  the  following  equations  : 

63.  .3x-  .02  -  .003  x  =  .7-  .06  a-  -  .006. 

64.  .001  x  -  .32  =  .09  x  -  .2  x  -  .653. 

65.  .3  (1.2  X-5)=U  +  .05  x. 

66.  .7  (x  +  .13)  =  .03  (4x-  .1)  +  .5. 

67.  3.3a;-       \         =.la;  +  9.9. 

.5 

2-3a       5a:    _  2x-  3  _  a-  -  2      0  7 
"T5—  +  L25~     ~9~       "378     ^  "  9  ' 

178.     To  prove  that  a  simple  equation  ran  have  but  one  root. 

We  have  soon  that  every  simple  equation  can  be  reduced  to 
the  form  x  =  a. 

Suppose,  if  possible,  that  a  simple  equation  can  have  two 
roots,  and  that  ->\  and  r.,  are  the  roots  of  the  equation  x  =  a. 
Then  (Art.  168), 

rx  —  a, 

r2  =  a. 

Hence,  tx  =  r2\  that  is,  the  two  supposed  roots  are  identical. 
Therefore  a  simple  equation  can  have  but  one  root. 


PROBLEMS.  103 


XIII.  — PROBLEMS 

LEADING  TO  SIMPLE  EQUATIONS   CONTAINING  ONE 
UNKNOWN  QUANTITY. 

179.  A  Problem  is  a  question  proposed  for  solution. 

180.  The  Solution  of  a  problem  by  Algebra  consists  of 
two  distinct  parts  : 

1.  The  Statement,  or  the  process  of  expressing  the  condi- 
tions of  the  problem  in  algebraic  language,  by  one  or  more 
equations. 

2.  The  Solution  of  the  resulting  equation  or  equations,  or 
the  process  of  determining  from  them  the  values  of  the  un- 
known quantities. 

The  statement  of  a  problem  often  includes  a  consideration 
of  ratio  and  proportion  (Art.  21). 

181.  Ratio  is  the  relation,  with  respect  to  magnitude, 
which  one  quantity  bears  to  another  of  the  same  kind,  and  is 
the  result  arising  from  the  division  of  one  quantity  by  the 
other. 

A  Proportion  is  an  equality  of  ratios. 
Thus, 

a  :  b,  or  -  ,  indicates  the  ratio  of  a  to  b. 

a  :  b  =  c  :  d,  is  a  proportion,  indicating  that  the  ratio  of  a 
to  b,  is  equal  to  the  ratio  of  c  to  d. 

In  a  proportion  the  relation  of  the  terms  is  such  that  the 
product  of  the  first  and  fourth  is  equal  to  the  product  of  the 
second  and  third. 

ct       c 
For,  a  :  b  =  c :  <%  is  the  same  as  j  —  -,  which,  by  clearing  of 

fractions,  gives  ad  =  b  c. 


1 04  ALGEBRA. 

182.  For  tlie  statement  of  a  problem  no  general  rule  can 
be  given ;  much  must  depend  on  the  skill  and  ingenuity  of  the 
operator.  We  will  give  a  few  suggestions,  however,  which 
will  be  found  useful  : 

1.  Express  the  unknoivn  quantity,  <>r  one  of  the  unknoivn 
quantities,  by  taw  of  the  final  letter*  of  the  alphabet. 

1'.  From  tlie  given  conditions,  find  expressions  for  the  other 
unknown  quantities,  if  any,  in  the  problem. 

3.  Form  on  equation,  by  indicating  the  operations  necessary 
to  verify  the  values  of  -the  unknown  quantities,  were  they 
already  known. 

4.  Determine  the  value  of  the  unknown  quantity  in  the 
equation  th  US  formed. 

Note.  Problems  which  involve  several  unknown  quantities  may  often 
be  solved  by  representing  one  of  them  only  by  a  single  unknown  letter. 

1.  What  number  is  that  to  which  if  four  sevenths  of  itself 
be  added,  the  sum  w.ill  equal  twice  the  number,  diminished  by 
27? 

Let  x  =  the  number. 

4  x 
Then  -=—  =  four  sevenths  of  it, 

and  2x  =  twice  it. 

4  x 

By  the  conditions,  x  -\ — —  =  2  x  —  27 

Solving  this  equation,         x  =  63,  the  number  required. 

2.  Divide  144  into  two  parts  whose  difference  is  30. 

Let  x  =  one  part. 

Tli en.  144  —  x  =  the  other  part. 

By  the  conditions,  x  —  (144  —  x)  =  30 

Solving  this  equation,  x  =  87,  one  part. 

144  — #=  ~>7,  the  other  part. 


PROBLEMS.  105 

3.  A  is  three  times  as  old  as  B  ;  and  eight  years  ago  he  was 
seven  times  as  old  as  B.     What  are  their  ages  at  present  ? 

Let  x  =  B's  age. 

Then,  3  x  —  A's  age. 

Now,  x  —  8  =  B's  age,  eight  years  ago, 

and  3  x  —  8  =  A's  age,  eight  years  ago. 
By  the  conditions,     3x  —  8  =  7  (x  —  8) 

Whence,  x  =  12,  B's  age, 

and,  3  x  =  36,  A's  age. 

4.  A  can  do  a  piece  of  work  in  8  days,  which  B  can  perform 
in  10  days.  In  how  many  days  can  it  he  done  hy  both  work- 
ing together  ? 

Let  x  =  the  number  of  days  required. 

Then,  -  =  what  both  can  do  in  one  day. 

Also,  —  =  what  A  can  do  in  one  day, 

o 

and  j-  =  what  B  can  do  in  one  day. 

Since  the  sum  of  what  each  separately  can  do  in  one  day  is 
equal  to  what  both  can  do  together  in  one  day, 

i     JL  -i 

8+10_x 

Whence,  x  =  4f ,  number  of  days  required. 

5.  A  man  has  $  3.64  in  dimes,  half-dimes,  and  cents.  He 
has  7  times  as  many  cents  as  half:dimes,  and  one  fourth  as 
many  half-dimes  as  dimes.     How  many  has  he  of  each  ? 


106  ALGEBRA. 

Let  x  =  the  number  of  dimes. 

x 

Then,  -   =  the  number  of  half-dimes, 

4 

7  x 
and  — r—  =  the  number  of  cents. 

4 

Now,  10  x  =  the  value  of  the  dimes  in  cents, 

and  —j—  ==  the  value  of  the  half-dimes  in  cents. 

4 

By  the  conditions,  10  x  -\ ; — | -—  =  364 

J  '44 

Whence,  x  =  28,  number  of  dimes, 

x 

-  =    7,  number  of  half-dimes, 

4 

— t—  =  49,  number  of  cents. 


6.  Two  pieces  of  cloth  were  purchased  at  the  same  price  per 
yard ;  but  as  they  were  of  different  lengths,  the  one  cost  $  5 
and  the  other  $  6.50.  If  each  had  been  10  yards  longer,  their 
lengths  would  have  been  as  5  to  6.  Required  the  length  of 
each  piece. 

Since  the  price  of  each  per  yard  is  the  same,  the  lengths  of 
the  two  pieces  must  be  in  the  ratio  of  their  prices,  that  is,  as  5 
to  6h,  or  as  10  to  13.     Therefore, 

Let  10  x  =  the  length  of  the  first  piece  in  yards, 

and  13  x  =  the  length  of  the  second  piece  in  yards. 

By  the  conditions,  10  x  +  10  :  13  x  +  10  =  5  :  6 

or  (Art.  181),     6  (10  x  +  10)  =  5  (13  x  +  10) 

"Whence,  x  =  2. 

Then,  10  x  =  20,  length  of  first  piece, 

and  13  x  =  26,  length  of  second  piece. 


PROBLEMS.  107 

7.  The  second  digit  of  a  number  exceeds  the  first  by  2 ;  and 
if  the  number,  increased  by  6,  be  divided  by  the  sum  of  the 
digits,  the  quotient  is  5.     Required  the  number. 

Let  x  =  the  first  digit. 

Then,         x  +  2  =  the  second. 

Since  the  number  is  equal  to  10  times  the  first  digit,  plus 
the  second, 

10  x  +  x  +  2,  or  11  x  +  2  =  the  number. 

11  x  _|_  2  +  6 

By  the  conditions,        ■ ^—  =  5 

J  x  +  x  +  2 

Whence,  x  =  2,  the  first  digit, 

and  x  +  2  =  4,  the  second  digit. 

Therefore  the  number  is  24. 

8.  Two  persons,  A  and  B,  63  miles  apart,  set  out  at  the 
same  time  and  travel  towards  each  other.  A  travels  4  miles 
an  hour,  and  B  3  miles.  What  distance  will  each  have  trav- 
elled when  they  meet  ? 

Let  x  =  the  distance  A  travels. 

Then,    63  —  x  =  the  distance  B  travels. 

x 

-  =  the  time  A  takes  to  travel  x  miles, 

and      — — —  =  the  time  B  takes  to  travel  63  —  x  miles. 
o 

By  the  conditions  of  the  problem,  these  times  are  equal ; 
x      63  —  x 

4=^r- 

Whence,  x  =  36,  A's  distance, 

and  63  —  x  =  27,  B's  distance. 


108  ALGEBRA. 

9.    At  what  time  between  3  and  4  o'clock  are  the  hands  of  a 
watch  opposite  to  each  other  ? 

Let  0  M  represent  the  position  of 
the  minute-hand  at  3  o'clock,  and  0  H 
the  position,  of  the  hour-hand  at  the 
same  time. 
\jj  Let  0  M1  represent  the  position  of 
Ih'  the  minute-hand  when  it  is  opposite 
to  the  hour-hand,  and  0  H1  the  po- 
sition of  the  hour-hand  at  the  same 
time. 

Let  x  =  the  arc  M H  H'  M',  the  space  over  which  the  min- 
ute-hand has  moved  since  3  o'clock. 

x 
Then,  ^  =  the  arc  H H',  the  space  over  which  the  hour- 

hand  has  moved  since  3  o'clock. 

Also,    the  arc  MH=  15  minute  spaces, 

and    the  arc  H'  M1  =  30  minute  spaces. 

Now,  arc  M H H1  M  =  arc  MH+  arc  H H<  +  arc  H<  M, 


x 
or,  x  =  15  +  j^  +  30 

Solving  this  equation,      x  =  49 ^  minute  spaces. 
That  is,  the  time  is  49-^  minutes  after  3  o'clock. 

PROBLEMS. 

10.  My  horse  and  chaise  are  worth  $  336 ;  but  the  horse  is 
worth  twice  as  much  as  the  chaise.  Required  the  value  of 
each. 

11.  What  number  is  that  from  which  if  7  be  subtracted,  one 
sixth  of  the  remainder  will  be  5? 

12.  What  two  numbers  are  those  whose  difference  is  3,  and 
the  difference  of  whose  squares  is  51  ? 


PROBLEMS.  109 

13.  Divide  20  into  two  such  parts  that  3  times  one  part  may 
be  equal  to  one  third  of  the  other. 

14.  Divide  100  into  two  parts  whose  difference  is  17. 

15.  A  is  twice  as  old  as  B,  and  10  years  ago  he  was  3  times 
as  old.     What  are  their  ages  ? 

16.  A  is  four  times  as  old  as  B ;  in  thirty  years  he  will  be 
only  twice  as  old  as  B.     What  are  their  ages  ? 

17.  A  can  do  a  piece  of  work  in  3  days,  and  B  can  do  the 
same  in  5  days.  In  how  many  days  can  it  he  done  by  both 
working  together  ? 

18.  A  can  do  a  piece  of  work  in  3§  hours,  which  B  can  do 
in  2|  hours,  and  C  in  2i  hours.  In  how  many  hours  can  it  be 
done  by  all  working  together  ? 

19.  A  and  B  can  do  a  piece  of  work  together  in  7  days, 
which  A  alone  can  do  in  10  days.  In  what  time  could  B  alone 
do  it  ? 

20.  The  first  digit  of  a  certain  number  exceeds  the  second 
by  4;  and  when  the  number  is  divided  by  the  sum  of  the 
digits,  the  quotient  is  7.      What  is  the  number  '.' 

21.  The  second  digit  of  a  certain  number  exceeds  the  first 
by  3;  and  if  the  number,  diminished  by  9,  be  divided  by  the 
difference  of  the  digits,  the  quotient  is  9.  What  is  the 
number  ? 

22.  A  drover  has  a  lot  of  oxen  and  cows,  for  which  he  gave 
$  1428.  For  the  oxen  he  gave  $  55  each,  and  for  the  cows  $  32 
each  ;  and  he  had  twice  as  many  cows  as  oxen.  Required  the 
number  of  each. 

23.  A  gentleman,  at  his  decease,  left  an  estate  of  $1872  for 
his  wife,  three  sons,  and  two  daughters.  His  wife  was  to  re- 
ceive three  times  as  much  as  either  of  her  daughters,  and  each 
son  to  receive  one  half  as  much  as  each  of  the  daughters.  Re- 
quired the  sum  that  each  received. 


HO  ALGEBRA. 

24.  A  laborer  agreed  to  serve  for  36  days  on  these  condi- 
tions, that  for  every  day  he  worked  he  was  to  receive  $1.25, 
but  for  every  day  lie  was  absent  he  was  to  forfeit  *  0.50.  At 
the  end  of  the  time  he  received  $  17.     It  is  required  to  find 

,  how  many  days  he  labored,  and  how  many  days  he  was  absent. 

25.  A  man,  being  asked  the  value  of  his  horse  and  saddle, 
replied  that  his  horse  was  worth  $114  more  than  his  saddle, 
and  that  g  the  value  of  the  horse  was  7  times  the  value  of  the 
saddle.     What  was  the  value  of  each  ? 

26.  In  a  garrison  of  2744  men,  there  are  2  cavalry  soldiers 
to  25  infantry,  and  half  as  many  artillery  as  cavalry.  Re- 
quired the  number  of  each. 

27.  The  stones  which  pave  a  square  court  would  just  cover 
a  rectangular  area,  whose  length  is  6  yards  longer,  and  breadth 
4  yards  shorter,  than  the  side  of  the  square.  Find  the  area  of 
the  court. 

28.  A  person  has  travelled  altogether  3036  miles,  of  which 
he  has  gone  7  miles  by  water  to  4  on  foot,  and  5  by  water  to 
2  on  horseback.  How  many  miles  did  he  travel  in  each 
manner  ? 

29.  A  certain  man  added  to  his  estate  ^  its  value,  and  then 
lost  $  760 ;  but  afterwards,  having  gained  $  600,  his  property 
then  amounted  to  $  2000.  What  was  the  value  of  his  estate  at 
first? 

30.  A  capitalist  invested  §  of  a  certain  sum  of  money  in 
government  bonds  paying  5  per  cent  interest,  and  the  re- 
mainder in  bonds  paying  6  per  cent  ;  and  found  the  interest 
of  the  whole  per  annum  to  be  $180.  Required  the  amount  of 
each  kind  of  bonds. 

31.  A  woman  sells  half  an  egg  more  than  halt  her  eggs. 
Again  she  sells  half  an  egg  more  than  half  her  remaining 
eggs.  A  third  time  she  does  the  same;  and  now  she  has  sold 
all  her  eggs.     How  many  had  she  at  first  ? 


PROBLEMS.  HI 

32.  What  number  is  that,  the  treble  of  which,  increased  by 
12,  shall  as  much  exceed  54,  as  that  treble  is  less  than  144  ? 

33.  A  ashed  B  how  much  money  he  had.  He  replied,  "  If 
I  had  5  times  the  sum  I  now  possess,  I  could  lend  you  $  60, 
and  then  i  of  the  remainder  would  be  equal  to  h  the  dollars  I 
now  have."     Required  the  sum  B  had. 

34.  A,  B,  and  C  found  a  purse  of  money,  and  it  was  mutu- 
ally agreed  that  A  should  receive  $  15  less  than  one  half,  that 
B  should  have  $13  more  than  one  quarter;  and  that  C  should 
have  the  remainder,  which  was  $  27.  How  many  dollars  did 
the  purse  contain? 

35.  A  number  consists  of  6  digits,  of  which  the  last  to  the 
left  hand  is  1.  If  tins  number  is  altered  by  removing  the  1 
and  putting  it  in  the  units'  place,  the  new  number  is  three 
times  as  great  as  the  original  one.     Find  the  number. 

36.  A  prize  of  $  1000  is  to  be  divided  between  A  and  B,  so 
that  their  shares  may  be  in  the  ratio  of  7  to  8.  Required  the 
share  of  each. 

37.  A  man  has  $  4.04  in  dollars,  dimes,  and  cents.  He  has 
one  fifth  as  many  cents  as  dimes,  and  twice  as  many  cents  as 
dollars.     How  many  has  he  of  each  ? 

38.  I  bought  a  picture  at  a  certain  price,  and  paid  the  same 
price  for  a  frame ;  if  the  frame  had  cost  $  1.00  less,  and  the 
picture  $  0.75  more,  the  price  of  the  frame  would  have  been 
only  half  that  of  the  picture.  Required  the  cost  of  the 
picture. 

39.  A  gentleman  gave  in  charity  $  46 ;  a  part  in  equal  por- 
tions to  5  men,  and  the  rest  in  equal  portions  to  7  women. 
Now,  a  man  and  a  woman  had  between  them  $8.  What 
was  given  to  the  men,  and  what  to  the  women  ? 

40.  Separate  41  into  two  such  parts,  that  one  divided  by 
the  other  may  give  1  as  a  quotient  and  5  as  a  remainder. 


112  ALGEBRA. 

41.  A  vessel  can  be  emptied  by  three  taps ;  by  the  first 
alone  it  could  be  emptied  in  80  minutes,  by  the  second  in  200 
minutes,  and  by  the  third  in  5  hours.  In  what  time  will  it  be 
emptied  if  all  the  taps  be  opened  ? 

42.  A  general  arranging  his  troops  in  the  form  of  a  solid 
square,  finds  he  has  21  men  over;  but,  attempting  to  add 
1  man  to  each  side  of  the  square,  finds  he  wants  200  men  to 
fill  up  the  square.  Required  the  number  of  men  on  a  side  at 
first,  and  the  whole  number  of  troops. 

43.  At  what  time  between  7  and  8  are  the  hands  of  a  watch 
opposite  to  each  other  ? 

44.  At  what  time  between  2  and  3  are  the  hands  of  a  watch 
opposite  to  each  other? 

45.  At  what  time  between  5  and  6  are  the  hands  of  a  watch 
together  ? 

46.  Divide  43  into  two  such  parts  that  one  of  them  shall  be 
3  times  as  much  above  20  as  the  other  wants  of  17. 

47.  Gold  is  19}  times  as  heavy  as  water,  and  silver  10i 
times.  A  mixed  mass  weighs  4160  ounces,  and  displaces  250 
ounces  of  water.  What  proportions  of  gold  and  silver  does  it 
contain  ? 

48.  A  gentleman  let  a  certain  sum  of  money  for  3  years  at 
5  per  cent  compound  interest ;  that  is,  at  the  end  of  each  year 
there  was  added  J,,  to  the  sum  due.  At  the  end  of  the  third 
year  there  was  due  him  .$2315.25.     Required  the  sum  let. 

49.  A  merchant  ha!s  grain  worth  9  shillings  per  bushel3  and 
other  grain  worth  1.'!  shillings  per  bushel,  in  what  proportion 
must  he  mix  40  bushels,  so  that  he  may  sell  the  mixture  at 
10  shillings  per  bushel '.' 

50.  A  alone  could  perform  a  piece  of  work  in  L2  hours;  A 
and  C  together  could  do  it  in  5  hours;  and  C's  work  is  §  of 
B's.  Now.  the  work  has  to  be  completed  by  noon.  A  begins 
work  at  5  o'clock  in  the  morning;  at  what  hour  can  he  he 
relieved  by  B  and  ( '.  ami  the  work-  he  just  finished  in  time'.' 


SIMPLE  EQUATIONS.  H3 

51.  A  merchant  possesses  $5120,  but  at  the  beginning  of 
each  year  he  sets  aside  a  fixed  sum  for  family  expenses.  His 
business  increases  his  capital  employed  therein  annually  at  the 
rate  of  25  per  cent.  At  the  end  of  four  years  he  finds  that  his 
capita]  is  reduced  to  $3275.     What  are  his  annual  expenses? 

52.  At  what  times  between  7  and  8  o'clock  are  the  hands  of 
a  watch  at  right  angles  to  each  other  ? 

53.  At  what  time  between  4  and  5  o'clock  is  the  minute- 
hand  of  a  watch  exactly  five  minutes  in  advance  of  the  hour- 
hand  ? 

54.  A  person  has  11^  hours  at  his  disposal ;  how  far  may 
he  ride  in  a  coach  which  travels  5  miles  an  hour,  so  as  to  re- 
turn home  in  time,  walking  back  at  the  rate  of  oh  miles  an 
hour  ? 

55.  A  fox  is  pursued  by  a  greyhound,  and  is  60  of  her  own 
leaps  before  him.  The  fox  makes  9  leaps  while  the  greyhound 
makes  but  6  ;  but  the  latter  in  3  leaps  goes  as  far  as  the  former 
in  7.  How  many  leaps  does  each  make  before  the  greyhound 
catches  the  fox  ? 

56.  A  clock  has  an  hour-hand,  a  minute-hand,  and  a  second- 
hand, all  turning  on  the  same  centre.  At  12  o'clock  all  the 
hands  are  together,  and  point  at  12.  How  long  will  it  be 
before  the  minute-hand  will  be  between  the  other  two  hands, 
and  equally  distant  from  each  ? 


XIV.  —  SIMPLE     EQUATIONS 

CONTAINING  TWO   UNKNOWN  QUANTITIES. 

183.  If  we  have  a  simple  equation  containing  two  unknown 
quantities,  as  3  x  —  4  y  =  2,  we  cannot  determine  definitely 
the  values  of  x  and  y ;  because,  for  every  value  which  we  give 
to  one  of  the  unknown  quantities,  we  can  find  a  corresponding 


114  ALGEBRA. 

value  for  the  other,  and  thus  find  any  number  of  pairs  of  values 
which  will  satisfy  the  given  equation. 

Thus,  if  we  put  x  =  G,       then  18  —  4  y  =  2,  or  y  =  4  ; 

t  if  we  put  x  =  —  2,  then  —  6  —  4  y  =  2,  or  y  =  —  2  ; 
if  we  put  a;  =  1,       then  3  —  4  ?/  =  2,  or  y  =  £ ;  etc. 

And  any  of  the  pairs  of  values  <       "  ,  I,  ■!  n  i,  ■<       ~  ,    , 

etc.,  will  satisfy  the  given  equation. 

If  we  have  another  equation  of  the  same  kind,  as  5x  +  7?/=17, 
we  can  find  any  number  of  pairs  of  values  which  will  satisfy 
this  equation  also. 

Now  suppose  we  are  required  to  determine  a  pair  of  values 
which  will  satisfy  both  equations.  We  shall  find  but  one  pair 
of  values  in  this  case.  For,  multiply  the  first  equation  by  5 ; 
thus, 

15  x -20  y  =  10; 

and  multiply  the  second  equation  by  3 ;  thus, 

15  x  +  21  y  =  51. 

Subtracting    the   first   of    these   equations    from    the    second 
(Art.  44),  we  have 

41  y  =  41, 

or,  p  =  l. 

In  the  first  given   equation  put  y  =  l;   then  3  x  —  4  =  2,  or 

3  x  =  6  ;  whence,  x  =  2.     The  pair  of  values  \     ~-i\  satisfies 

both  the  given  equations;  and  no  other  pair  of  values  can  be 
found  which  will  satisfy  both. 

184.  Simultaneous  Equations  are  such  as  are  satisfied  by 
the  same  values  of  their  unknown  quantities. 

185.  Independent  Equations  are  such  as  cannot  be  made 
to  assume  the  same  form. 


SIMPLE  EQUATION'S.  115 

186.  It  is  evident,  from  Art.  183,  that  two  unknown 
quantities  require  for  their  determination  two  independent, 
simultaneous  equations.  When  two  such  equations  are  given, 
it  is  our  object  to  obtain  from  them  a  single  equation  contain- 
ing but  one  unknown  quantity.  The  value  of  that  unknown 
quantity  may  then  be  found;  and  by  substituting  it  in  either 
of  the  given  equations  we  can  find,  as  in  Art.  183,  the  value  of 
the  other. 

ELIMINATION". 

187.  Elimination  is  the  process  of  combining  simultaneous 
equations  so  as  to  obtain  from  them  a  single  equation  contain- 
ing but  one  unknown  quantity. 

There  are  four  principal  methods  of  elimination :  by  Addi- 
tion or  Subtraction,  by  Substitution,  by  Comparison,  and  by 
Undetermined  Multipliers. 

CASE    I. 

188.  Ellin  {nation  by  Addition  or  Subtraction. 

1.  Given  ox  —  3  y  =  19,  and  7  x  +  4  y  —  2,  to  find  the 
values  of  x  and  y. 

Multiplying  the  first  equation  by  4,        20  x  —  12  y  =  76 
Multiplying  the  second  equation  by  3,   21  x  +  12  y  =    6 

Adding  these  equations,  41  x  =  82 

Whence,  x  =  2. 

Substituting  this  value  in  the  first  given  equation, 

10-3y  =  19 
-3y  =  9 
y  =  -3. 

We  might  have  solved  the  equations  as  follows : 

Multiplying  the  first  by  7,  35  x  -  21  y  =  133    (1) 

Multiplying  the  second  by  5,  35  x  +  20  y  =    10    (2) 

Subtracting  (2)  from  (1),  —  41  y  =  123 

2/  =  -3. 


116  ALGEBRA. 

Substituting  this  value  of  y  in  the  first  given  equation, 

5  x  +  9  =  19 
5  a- =  10 
x  =  2. 

The  first  of  these  methods  is  elimination  by  addition  ;  the 
second,  elimination  by  subtraction. 

RULE. 

Multiply  the  given  equations,  if  necessary,  by  such  numbers 
or  quantities  as  will  make  the  coefficient  of  one  of  the  unknown 
quantities  the  same  in  the  two  resulting  equations.  Then,  if 
the  signs  of  the  terms  having  the  same  coefficient  arc  alike, 
subtract  one  equation  from  the  other  ■  if  unlike,  add  the  two 
equations. 

This  method  of  elimination  is  usually  the  best  in  practice. 

CASE    II. 
189.     Elimination  by  Substitution. 
Taking  the  same  equations  as  before, 

5  x  —  3  y  =  19  (1) 

7x  +  4t/=    2  (2) 

Transposing  the  term  7  x  in  (2),       4  y  =  2  —  7  x 

2 7  x 

Dividing  by  4,  y  =  — _  (3) 

Substituting  this  value  of  y  in  (1), 

5*_3  (^=^)  =19 

Performing  the  operations  indicated, 


SIMPLE   EQUATIONS.  117 

Clearing  of  fractions,       20  x  —  (6  —  21  x)  =  76 
or,  20  x  -  6  +  21  x  =  76 

Transposing,  and  uniting  terms,  41  x  =  82 

Whence,  x  =  2. 

2  —  14 

Substituting  this  value  in  (3),  y  =  — j —  =  —  3. 

•     RULE. 

7v//r/  £/ie  7v//»e  o/o/^e  o/  £Ae  unknown  quantities  in  terms 
of  the  other,  from  cither  of  the  given  equations;  and  substi- 
tute this  value  for  that  quantity  in  the  other  equation. 

This  method  is  advantageous  when  either  of  the  unknown 
quantities  has  1  for  its  coefficient. 

CASE    III. 
190.     Elimination  by  Comparison. 

Taking  the  same  equations  as  before, 

5  x  -  3  y  =  19  (1) 

7x  +  ±y=   2  (2) 

Transposing  the  term  —  3  y  in  (1),        5  x  =  3  y  +  19 

3//+ 19  ,,. 

or,  a  =  — g (3) 

Transposing  the  term  4  y  in  (2),  7  a;  =  2  —  4  y 

or,  a:  = — 

Placing  these  two  values  of  x  equal  to  each  other  (Art.  44), 

3//  +  19_2-4y 

5  7 

Clearing  of  fractions,       21  y  +  133  =  10  —  20  y 


118  ALGEBRA. 

Transposing,  and  uniting  terms,      41  y  =  —  123 
Whence,  y  —  —  3. 

—  9  +  19 

Substituting  this  value  in  (3),  x  = p 

o 


RULE. 

Find  the  value  of  the  same  unknown  quantity  in  terms  of 
the  other,  from  each  of  the  given  equations  :  ami  form  a  new 
equation  by  placing  these  values  equal  to  each  other. 

CASE    IV. 

191.     Elimination  by  Undetermined  Multipliers. 

An  Undetermined  Multiplier  is  a  factor,  at  first  undeter- 
mined, but  to  which  a  convenient  value  is  assigned  in  the 
course  of  the  operation. 

Taking  the  same  equations  as  before, 

5x-3y  =  19  (1) 

7  x  +  4  y  =   2  (2) 

Multiplying  (1)  by  m,     5  m  x  —  3  m  y  =  19  m  (3) 

Subtracting  (3)  from  (2), 

7  x  —  5mx  +  iy  +  3m  y  —  2  —  19  m 
Factoring,     x  (7  —  5  m)  +  y  (4  +  3  m)  =2  —  19  m  (4) 

Now,  let  the  coefficient  of  y,   4  +  3  m  =  0  ;  then  3  m  =  —  4, 

4 

or  m  =  —  Kj  substituting  this  value  of  m  in  (4), 
o 

/_      20\     „     76 

n7+ir)  =  2+3 

Clearing  of  fractions,       x  (21  +  20)  =  6  +  70 

41  x  =  82 
x  =  2. 


SIMPLE   EQUATIONS.  119 

Substituting  this  value  in  (2),     14  +  4  y  =  2 

4y  =  -12 
y  =  -3. 

We  might  liave  let  the  coefficient  of  x  in  (4),  7  —  5m  =  0; 

7 
then  m  would  have  been  ■=  ;  substituting  this  value  of  m  in  (4), 

o 


y(±+       )=2-- 


Clearing  of  fractions,     y  (20  +  21)  =  10  -  133 

41  y  =  -  123 
y  =  -3. 

Instead  of  subtracting  (3)  from  (2),  we  migbt  have  added 
them  and  obtained  the  same  results.  Also,  in  the  first  place, 
we  might  have  multiplied  (2)  by  m,  and  either  added  the  re- 
sult to,  or  subtracted  it  from,  (1). 


RULE. 

Multiply  one  of  the  given  equations  by  the  undetermined 
quantity,  m  ;  and  add  the  result  to,  or  subtract  it  from,  the 
other  given  equation. 

In  the  resulting  equation,  factored  with  reference  to  the 
unknown  quantifies,  place  the  coefficient  of  one  of  the  un- 
known quantities  equal  to  zero,  and  find  the  value  of  ra. 
Substitute  tins  value  of  m  In  the  equation,  and  the  result  trill 
be  a  simple  equation  containing  but  one  unknown  quantity. 

This  method  is  advantageous  in  the  solution  of  literal 
equations. 

2.    Solve  the  equations. 

ax  +  b  y  —  c  (1) 

afx  +  b'y  =  c'  (2) 


120  ALGEBRA. 

Multiplying  (1)  by  m,  a  m  x  +  b  m  y  =  c  m  (3) 

Add  (2)  and  (3),       a'  x  +  a  m  x  +  V  y  +  b  m  y  =  d +  cm 
Factoring,  x  (a'  +  a  vi)  +  y  (b1  +  b  m)  =  d  +  c  m     (4) 

In  (4),  put  the  coefficient  of  y,  V  +  b  in,  equal  to  zero. 

Then,  b  m  —  —  V  ;  whence,  m  =  —  -. 

b 

Substituting  this  value  of  m  in  (4), 


a  V  \        ,      c  b 


x  \a' —  )  =  d 


Clearing  of  fractions,     x  (a'  b  —  ab')  =b  d  —  b'  e 

1,  (j  _  y  c 

Whence,  x  =  — — . 

a'  b  —  ab' 

In  (4),  put  the  coefficient  of  x,  a'  +  a  m,  equal  to  zero. 
Then,  a  m  =  —  a ' ;  whence,  m  =  ■ . 


a 


Substituting  this  value  of  m  in  (4), 


a'b\  a'e 

y\h — '-)  —  c  — 


a  '  a 


Clearing  of  fractions,      y  (ab1  —  a1  b)  =  ad  —  a'  c 

a  d  —  a'  c 

Wh  ence,  y  =  —rl — . 

ab'  —  a'  b 

Before  applying  either  of  the  preceding  methods  of  elimina- 
tion, the  given  equations  should  be  reduced  to  their  simplest 
forms. 

EXAMPLES. 

192.  Solve,  by  whichever  method  may  be  most  advanta- 
geous, the  following  equations : 

3.  3cc  +  72/  =  33;  2x  +  ±y  =  20. 

4.  7x  +  2y  =  31;  3  z  ~  4  ?/ =  23. 

5.  6x-3y  =  27;  4;r-6y  =  -2. 


SIMPLE   EQUATIONS.  121 

6.  7  x  +  3y  =  -50;  2y-5x  =  U. 

7.  8y  +  12 .£  =  116  ;  2x  —  y  =  3. 

8.  11  x  +  3  y  =  -  124  ;  2  x  -  6  y  =  56. 

9.  9a;  +  4?/  =  22;27/  +  3cc  =  14. 

10.  ^+^f-8;  -8x  +  2y  =  -S0. 

11.  7z-2y  =  G;2a;  +  22/  =  -24. 

,«-.-  ^         ,.„     2^;       11?/  5 

12.  11  y  +  G  a;  =  115  ;  — —  =  —  ■= . 

13.  |a}  +  |y  =  ^;  10*-12y  =  -62. 

5        7 

14.  _7a;  +  47/  =  -113;  £  +  -?,  =  -. 

15>    2~3-°'  4  +  G-b- 

16.  ^-y  =  31;  a;  +  ^  =  33. 

17.  A^_^=_30;  aj  +  7y  =  119. 

<  3  a 

18.  rc  +  2//  =  .G;  1.7 x -y  =  . 58. 
3  ^      ?/     cc      2  ?/ 

19.  -tr-?-?1[gL=^.4y-8aJ  =  lL 

2  T 

»  +  3y  3        7  y - x 

'   2x-y  8'  2  +  x  +  2y  ~ 

21.  a  x  +  b  y  =  m ;  c  x  +  d  y=  n. 

22.  m  cc  +  ?i  ?/  =  r  ;  m'  cc  —  n'  y  =  /. 

no     xy  y      x 

aw  a      c 


122  ALGEBRA. 


x  v  1  x  y  1 

**   a  +  b^  a~b      a2-b2>  a-b^  a  +  b     a2-b2 

9*       X         19         Vm*.X         S         2V~X-Q7         X  +  V 

25l-2~12  =  4+8'  3~8 4T~~^ V 

2             2,3  2 

26.  + =  1;—  -  =  0. 

03  +  //  X-  —  V/  03  +  2/  ^  —  i7 

2x  +  y_17      2y  +  03     4      2  a?  -  y  _  2  y  -  a? 

4*.   x-     -g-    -12  4    "'  3  4    "-y       "    3      ' 

2  03      3.7/      x  +  2?/_        5x  — 6y 
28<   "3 5 4~~  ~°  4~"' 

03  ox  —  y  H       x 


29.    Solve  the  equations, 

6_3_ 
x     y~ 

8     15 

-  +  —  =  -1 

«     y 

Multiplying  the  first  equation  hy  5, 

30_15  =  2Q 

x      y  ' 

Adding  this  to  the  second  given  equation, 

"  =  19 

03 

Clearing  of  fractions,  38  =  19  03 

Whence  03  =  2. 

Substituting  this  value  in  the  first  given  equation, 

y 

Transposing,  =  1 

Whence,  y  =  —  3. 


SIMPLE  EQUATIONS.  123 

Solve  the  following  equations  : 


3     15    2     3 

-1. 

30. 

x      y      4 '  x      y 

31. 

12      18          42     15 

8 

17 

x        y            5  '    x 

y 

3 

32. 

11      7      3     4      8 
x       y      2 '  x      y 

-10. 

nn     a      b  c       d 

66.    -  +  -  =  m  ;   -+-  =  ». 
x      y  x      y 

fi  4  9  8 

34.   — +  f-  =  4a5;  -^ —  =  3a2-4Ja. 

<za;       6y  6a;       ay 

oc      m  ?i  n      m         ,o 

oO. 1 =zm  +  n;  — | =  iiv  +  n". 

nx       my  x       y 


XV.  — SIMPLE    EQUATIONS 

CONTAINING  MORE  THAN  TWO  UNKNOWN  QUANTITIES. 

193.     If  we  have   given   three   independent,   simultaneous 

equations,  containing  three  unknown  quantities,  we  may  coin- 
hine  two  of  them  hy  the  methods  of  elimination  explained  in 
the  last  chapter,  so  as  to  obtain  an  equation  containing  only 
two  unknown  quantities  ;  we  may  combine  the  third  equation 
with  either  of  the  two  former  in  the  same  way,  so  as  to  obtain 
another  equation  containing  the  same  two  unknown  quantities. 
Then  from  these  two  equations  containing  two  unknown  quan- 
tities we  may  derive,  as  in  the  last  chapter,  the  values  of  those 
unknown  quantities.  These  values  being  substituted  in  either 
of  the  given  equations,  the  value  of  the  third  unknown  quan- 
tity may  be  determined  from  the  resulting  equation. 

The  method   of  elimination  by  addition  or  subtraction  is 
usually  the  most  convenient. 


124  ALGEBRA. 

194.     1.    Solve  the  equations, 

8x-dy-7z  =  -36 

12  x—    y  —  3z  —     36 

6x-2y-    z  =      10 

Multiplying  the  first  by  3,  21  x  —  27  y  -  21  z  =  —  108  (1) 
Multiplying  the  second  by  2,  21  x  —  2  y  —  6  z  =  72  (2) 
Multiplying  the  third  by  4,  21  x  —    Sy  —    4tz  =        10      (3) 

Subtracting  (1)  from  (2), 

or, 

Subtracting  (3)  from  (2), 

or, 

Multiplying  (5)  by  3, 
Adding  (1)  and  (6), 

Substituting  this  value  in  (5), 
Substituting  the  values  of  y  and  z  in  the  third  given  equation, 

sc  =  4. 

In  the  same  manner,  if  we  have  given  n  feidependent, 
simultaneous  equations,  containing  n  unknown  quantities,  we 
may  combine  them  so  as  to  form  a  —  1  equations,  containing 
n  —  1  unknown  quantities.  These,  again,  may  be  combinnl 
so  as  to  form  n  —  2  equations,  containing  n  —  2  unknown 
quantities;  and  so  on  :  the  operation  being  continued  until  we 
finally  obtain  one  equation  containing  one  unknown  quantity. 

RULE. 

Multiply  the  given  equations,  if  necessary,  by  such  numbers 
or  quantities  as  trill  make  the  coeffirirnt  of  one  of  the  un- 
known quantities  the  same  in  the  resulting  equations.  Cont- 
inue these  equations  by  addition  <>r  subtraction,  so  as  to  form 


2oy  +  loz  =  ISO 

5y+    3z=   36 

(4) 

6y-   2z=   32 

3y-      z=    16 

(5) 

9y-   3z=   18 

(6) 

11//=   81 

y  =  6. 

z  =  2. 

SIMPLE  EQUATIONS.  125 

a  new  set  of  equations,  one  less  in  number  than  before,  and 
containing  one  less  unknown  quantity.  Continue  the  opera- 
tion with  these  new  equations  ;  and  so  on,  until  an  equation  is 
obtained  containing  nut-  unknown  quantity. 

Find  the  value  of  this  unknown  quantity.  By  substituting 
it  in  either  of  the  equations  containing  only  two  unknown 
quantities,  find  the  value  of  a  second  unknown  quantity.  By 
substituting  these  values  in  either  of  the  equations  containing 
three  unknown  quantities,  find  the  value  of  a  third  unknown 
quantity  j  and  so  on,  until  the  values  of  all  arc  found. 

Note.  This  rule  corresponds  only  with  the  method  of  elimination  by- 
addition  or  subtraction  ;  which,  however,  as  we  have  observed  before,  is 
the  best  in  practice. 

2.    Solve  the  equations, 

u  +  x  +  y  =  6 
u  +  x  -\-  z  =  9 

u  +  y  +  z  =  8 
x+y+z=7 

The  solution  may  here  be  abridged  by  the  artifice  of  assum- 
ing the  sum  of  the  four  unknown  .quantities  to  equal  an  auxil- 
iary quantity,  s.     Thus, 

Let  u  +  x  +  y  +  z  =  s. 

Then  we  may  write  the  four  given  equations  as  follows : 

s-z  =  6  (1) 

s-y=9  (2) 

s-x  =  S  (3) 

s-u=7  (4) 


Adding,  4  s  —  s  =  30 

Whence,  s  =  10. 

Substituting  the  value  of  s  in  (1),,(~),  (3),  and  (4),  we  obtain 

z  ■=  4,  y  =  1,  x  =  2,  and  u  =  3. 


126  ALGEBRA. 

EXAMPLES. 

Solve  the  following  equations  : 

3.  x  +  y+z  =  53;  x  +  2  y  +  3s  =  107;  x  +  3y  +  4s  =  137. 

4.  3x-y  —  2z  =  -23\    G  x  +  2  y  +  3  z  =  15; 

4x  +  3  y  —  z  =  —  G. 

5.  5x  —  3y+2z  =  41i  2x  +  y-z  =  l7;  5x  +  ±y-2z  =  3%. 

6.  7a;  +  47/-2  =  -50;    4x- 5 y-3  z  =  20; 

x  —  3y  —  4:Z  =  30. 

7.  3u  +  x  +  2y  —  z  =  22;    4x  —  y  +  3  s  =  35; 

4:u  +  3x-2y=19;    2 u  +  4y  +  2 2  =  46. 

8.  a;  +  ?/  =  2j    a2  +  s  =  3;    y  +  *  =  — 1. 

9.  ?/  +  s  =  a ;    aj  +  £  =  J ;    a;  +  y  =  c. 

10.   4a;-4y  =  a  +  4s;  6y  —  2a;  =  a  +  2s;  7s-y  =  fl  +  ^. 

11   2  +  3~4-~43'    3"I+2-°4'    4  +  2~3==~°°- 

12.  2a  +  22,  +  3  =  -17-2w;    ?/  +  3*=-2;    4aj.  +  *  =  13j 

^  +  3y  =  -14. 

13.  a  y  -\-  b  x  =  c;    c  x  +  a  z  =  b  ;    b  z  +  c  y  =  a. 


14    4 ?_  6 81     _2_     _3_      10    _7 

a:      y      2~=2~'    3^"+2l/~7ri"Z"2 

8       6       4        .. 

+  —  =  11. 


9  .'■      //      7 

.-      3         2  J_     J_  -1      A      A  -1 

°'    4x    "37/"      ;'  ""32/+  2z~    ;     2z  +  4:x~ 

16.   x  —  ay+  cr z  —  a8 ;  x  —  by+b'2z  =  b:i;  x  —  c y  +  c'2 z  =  c 


8 


PROBLEMS.  127 


17  y~z     x  +  % _ 1    x—y    x—z—n. 

2     ~     4     ~2'        5  6 


y+z       x  +  y 
4  2 


-4. 


18.   £^±1_(2_,)  =  0;    ^±-^  =  2a-cx 
a  c 

(«  +  f)2-ac(2  +  a;  +  s)  =  -y. 


XVI.  —  PROBLEMS 

LEADING    TO  SIMPLE    EQUATIONS    CONTAINING    MORE 
THAN   ONE   UNKNOWN   QUANTITY. 

195.  In  the  solution  of  problems  in  which  we  represent 
more  than  one  of  the  unknown  quantities  by  letters,  we  must 
obtain,  from  the  conditions  of  the  problem,  as  many  indepen- 
dent equations  as  there  are  unknown  quantities. 

1.  If  3  be  added  to  both  numerator  and  denominator  of  a 
certain  fraction,  its  value  is  f ;  and  if  2  be  subtracted  from 
both  numerator  and  denominator,  its  value  is  h-  Required 
the  fraction. 

Let 
and 

By  the  conditions, 


X- 

=  the 

numerator, 

y-- 

=  the  denominator, 

X 

+  3 

o 

V 

+  3" 

o 
O 

X 

o 

w 

1 

y 

-2 

2 

x  ■■ 

=  ~> 

y  =  12. 

Solving  these  equations, 

That  is,  the  fraction  is  Tv. 

2.  The  sum  of  the  digits  of  a  number  of  three  figures  is  13 ; 
if  the  number,  decreased  by  8,  be  divided  by  the  sum  of  the 
second  and  third  digits,  the  result,  is  25;  and  if  99  be  added 
to  the  number,  the  digits  will  be  inverted.     Find  the  number. 


128  ALGEBRA. 

Let  x  =  the  first  digit, 

y  =  the  second, 

and  z  —  the  third. 

Then,  100  x  +  10  y  +  z  =  the  nuniher, 
and     100  z  +  10  y  +  x  =  the  number  with  its  digits  inverted. 

By  the  conditions,  x  +  y  +  z  =  13 

100  x +  10  y  +  z-8 
—  Zo 

100  x  +  10  y  +  z  +  99  =  100  z  +  10  y  +  x 

Solving  these  equations,  x  =  2,  the  first  digit, 

y  =  8,  the  second, 


3,  the  third. 


That  is,  the  number  is  283. 


3.  A  crew  can  row  20  miles  in  2  hours  down  stream,  and 
12  miles  in  3  hours  against  the  stream.  Required  the  rate 
per  hour  of  the  current,  and  the  rate  per  hour  of  the  crew  in 
still  water. 

Let  x  =  rate  per  hour  of  the  crew  in  still  water, 

and  y  =  rate  jjer  hour  of  the  current. 

Then,  x  +  y=  rate  per  hour  rowing  down  stream, 
and      x  —  y  =  rate  per  hour  rowing  up  stream. 

Since  the  distance  divided  by  the  rate  gives  the  time,  we 
have  by  the  conditions, 

20         2 


x  +  y 
12 


=  3 


x—y 
Solving  these  equations,  x  =  7,  and  y  =  3. 


PROBLEMS.  129 

PROBLEMS. 

4.  A  says  to  B,  "  If  i  of  my  age  were  added  to  §  of  yours, 
the  sum  would  be  19^  years."  "  But,"  says  B,  "  if  §  of.  mine 
were  subtracted  from  J  of  youx-s,  the  remainder  would  be  18£ 
years."     Required  their  ages. 

5.  If  1  be  added  to  the  numerator  of  a  certain  fraction,  its 
value  is  1 ;  but  if  1  be  added  to  its  denominator,  its  value  is  £. 
What  is  the  fraction? 

6.  A  farmer  has  89  oxen  and  cows  ;  but,  having  sold  4  oxen 
and  20  cows,  found  he  then  had  7  more  oxen  than  cows.  Re- 
quired the  number  of  eacb  at  first. 

7.  A  says  to  B,  "  If  7  times  my  property  were  added  to  \  of 
yours,  the  sum  would  be  %  990."  B  replied,  "  If  7  times  my 
property  were  added  to  \  of  yours,  the  sum  would  be  $  510." 
Required  the  property  of  each. 

8.  If  \  of  A's  age  were  subtracted  from  B's  age,  and  5  years 
added  to  the  remainder,  the  sum  would  be  6  years  ;  and  if  4 
years  were  added  to  \  of  B's  age,  it  would  be  equal  to  fe  of  A's 
age.     Required  their  ages. 

9.  Divide  50  into  two  such  parts  that  %  of  the  larger  shall 
be  equal  to  §  of  the  smaller. 

10.  A  gentleman,  at  the  time  of  his  marriage,  found  that  his 
wife's  age  was  to  his  as  3  to  4 ;  but,  after  they  had  been  mar- 
ried 12  years,  her  age  was  to  his  as  5  to  6.  Required  their 
ages  at  the  time  of  their  marriage. 

11.  A  farmer  hired  a  laborer  for  10  days,  and  agreed  to  pay 
him  $  12  for  every  day  he  labored,  and  he  was  to  forfeit  %  8 
for  every  day  he  was  absent.  He  received  at  the  end  of  his 
time  %  40.  How  many  days  did  he  labor,  and  how  many  days 
was  he  absent  ? 

12.  A  gentleman  bought  a  horse  and  chaise  for  % 208,  and  i 
of  the  cost  of  the  chaise  was  equal  to  §  the  price  of  the  horse. 
What  was  the  price  of  each  ? 


130  ALGEBRA. 

13.  A  and  B  engaged  in  trade,  A  with  $  240,  and  B  with 
$96.  A  lost  twice  as  much  as  B;  and,  upon  settling  their 
accounts,  it  appeared  that  A  had  three  times  as  much  remain- 
ing as  B.     How  much  did  each  lose  ? 

14.  Two  men,  A  and  B,  agreed  to  dig  a  well  in  10  days; 
hut,  having  labored  together  4  days,  1!  agreed  to  finish  the 
job,  which  he  did  in  16  days.  How  long  would  it  have  taken 
A  to  dig  the  whole  well  ? 

15.  A  merchant  has  two  kinds  of  grain,  one  at  60  cents  per 
bushel,  and  the  other  at  90  cents  per  bushel,  of  which  he 
wishes  to  make  a  mixture  of  40  bushels  that  may  be  worth 
80  cents  per  bushel.  How  many  bushels  of  each  kind  must 
he  use  '.' 

16.  A  farmer  has  a  box  filled  with  wheat  and  rye;  seven 
times  the  bushels  of  wheat  are  3  bushels  more  than  four  times 
the  bushels  of  rye ;  and  the  quantity  of  wheat  is  to  -the  quan- 
tity of  rye  as  3  to  5.     Required  the  number  of  bushels  of  each. 

17.  My  income  and  assessed  taxes  together  amount  to  $  50. 
But  if  the  income  tax  be  increased  50  per  cent,  and  the  as- 
sessed tax  diminished  25  per  cent,  the  taxes  will  together 
amount  to  $  52.50.     Required  the  amount  of  each  tax. 

18.  A  and  B  entered  into  partnership,  and  gained  8200. 
Now  6  times  A's  accumulated  stock  (capital  and  profit)  was 
equal  to  5  times  B's  original  stock;  and  6  times  B's  profit 
exceeded  A's  original  stock  by  $200.  Required  the  original 
stock  of  each. 

19.  A  boy  at  a  fair  spent  his  money  for  oranges.  If  he  had 
got  five  more  for  his  money,  they  would  have  averaged  a  half- 
cent  less  ;  and  if  three  less,  a  half-cent  each  more.  J  low  many 
cents  did  he  spend,  and  how  many  oranges  did  he  get? 

20.  A  merchant  has  three  kinds  of  sugar.  He  can  sell  3 
lbs.  of  the  first  quality,  4  lbs.  of  the  second,  and  2  lbs.  of  the 
third,  for  60  cents;  or,  he  can  sell  4  lbs.  of  the  first  quality, 
1  lb.  of  the  second,  and  5  lbs.  of  the  third,  for  59  cents ;  or,  he 


PROBLEMS.  131 

can  sell  1  lb.  of  the  first  quality,  10  lbs.  of  the  second,  and  3 
lbs.  of  the  third,  for  90  cents.  Required  the  price  per  lb.  of 
each  quality. 

21.  A  gentleman's  two  horses,  with  their  harness,  cost  him 
$120.  The  value  of  the  poorer  horse,  with  the  harness,  was 
double  that  of  the  better  horse;  and  the  value  of  the  better 
horse,  with  the  harness,  was  triple  that  of  the  poorer  horse. 
What  was  the  value  of  each  ? 

22.  Find  three  numbers,  so  that  the  first  with  half  the  other 
two,  the  second  with  one  third  the  other  two,  and  the  third 
with  one  fourth  the  other  two,  shall  each  be  equal  to  34. 

23.  Find  a-  number  of  three  places,  of  which  the  digits  have 
equal  differences  in  their  order ;  and,  if  the  number  be  divided 
by  half  the  sum  of  the  digits,  the  quotient  will  be  41 ;  and,  if 
396  be  added  to  the  number,  the  digits  will  be  inverted. 

24.  There  are  four  men,  A,  B,  C,  and  D,  the  value  of  whose 
estates  is  $14,000;  twice  A's,  three  times  B's,  half  of  C's,  and 
one  fifth  of  J)'s,  is  $16,000;  As,  twice  B's,  twice  C's,  and  two 
fifths  of  D's,  is  $18,000;  and  half  of  A's,  with  one  third  of 
B's,  one  fourth  of  C's,  and  one  fifth  of  D's,  is  $  4000.  Re- 
quired the  property  of  each. 

25.  A  and  B  are  driving  their  turkeys  to  market.  A  says 
to  B,  "  Give  me  5  of  your  turkeys,  and  I  shall  have  as  many 
as  you."  B  replies,  "  Give  me  15  of  yours,  and  then  yours 
will  be  f  of  mine."     How  man}'  had  each  ? 

26.  A  says  to  B  and  C,  "  Give  me  half  of  your  money  and  I 
shall  have  $  55."  B  replies,  "  If  you  two  will  give  me  one 
third  of  yours,  I  shall  have  $  50."  But  C  says  to  A  and  B,  "  If 
I  had  one  fifth  of  your  money  I  should  have  $50."  Required 
the  sum  that  each  possessed. 

27.  A  gentleman  left  a  sum  of  money  to  be  divided  among 
his  four  sons,  so  that  the  share  of  the  eldest  was  \  of  the  sum 
of  the  shares  of  the  other  three,  the  share  of  the  second  J  of 
the  sum  of  the  other  three,  and  the  share  of  the  third  I  of  the 


132  ALGEBRA. 

sum  of  the  other  three ;  and  it  was  found  that  the  share  of  the 
eldest  exceeded  that  of  the  youngest  by  $  14.  "What  was  the 
whole  sum,  and  what  was  the  share  of  each  person? 

28.  If  I  were  to  enlarge  my  field  by  making  it  .")  rods  longer 
and  1  rods  wider,  its  area  would  be  increased  by  240  square 
rods;  but  if  I  were  to  make  its  length  4  rods  less,  and  its 
width  5  rods  less,  its  area  would  be  diminished  by  210  square 
rods.     Required  the  present  length,  width,  and  area. 

29.  A  boatman  can  row  down  stream,  a  distance  of  20  miles, 
and  back  again  in  10  hours;  and  he  finds  that  he  can  row  2 
miles  against  the  current  in  the  same  time  that  he  rows  3  miles 
with  it.     Required  the  time  in  going  and  in  returning. 

30.  A  and  B  can  perform  a  piece  of  work  in  0  days.  A  and 
C  in  8  days,  and  B  and  C  in  12  days.  In  how  many  "lays  can 
each  of  them  alone  perform  it  ? 

31.  A  person  possesses  a  capital  of  $30,000,  on  which  he 
gains  a  certain  rate  of  interest;  but  he  owes  $20,000,  for 
which  he  pays  interest  at  another  rate.  The  interest  which 
he  receives  is  greater  than  that  which  he  pays  by  $800.  A 
second  person  has  $35,000,  on  which  he  gains  the  second  rate 
of  interest ;  but  he  owes  $  24,000,  for  which  he  pays  the  first 
rate  of  interest.  The  sum  which  he  receives  is  greater  than 
that  which  he  pays  by  $  310.  What  are  the  two  rates  of  in- 
terest ? 

32.  A  man  rows  down  a  stream,  which  runs  at  the  rate  of 
o\  miles  per  hour,  for  a  certain  distance  in  1  hour  and  40  min- 
utes. In  returning  it  takes  him  6  hours  and  .'50  minutes  to 
arrive  at  a  point  2  miles  short  of  his  starting-place.  Find  the 
distance  lie  pulled  down  the  stream,  and  the  rate  of  his  pulling. 

33.  A  train  running  from  Boston  to  New  York  meets  with 
an  accident  which  causes  its  speed  to  be  reduced  to  ,1  of  what 
it  was  before,  and  it  is  in  consequence  5  hours  late.  If  the 
accident  bad  happened  60  miles  nearer  New  York,  the  train 
would  have  been  only  one  hour  late.  "What  was  the  rate  of 
the  train  before  the  accident  ? 


PROBLEMS.  133 

34.  A  and  B  run  a  mile.  A  gives  B  a  start  of  44  yards 
and  beats  him  by  51  seconds,  and  afterwards  gives  him  a  start 
of  1  minute  15  seconds  and  is  beaten  by  88  yards.  In  how 
many  minutes  can  each  run  a  mile  ? 

35.  A  merchant  has  two  casks,  each  containing  a  certain 
quantity  of  wine.  In  order  to  have  an  equal  quantity  in  each, 
he  pours  out  of  the  first  cask  into  the  second  as  much  as  the 
second  contained  at  first ;  then  he  pours  from  the  second  into 
the  first  as  much  as  was  left  in  the  first;  and  then  again  from 
the  first  into  the  second  as  much  as  was  left  in  the  second, 
when  there  are  found  to  be  1G  gallons  in  each  cask.  How 
many  gallons  did  each  cask  contain  at  first  ? 

36.  A  and  B  arc  building  a  fence  12G  feet  long;  after  three 
hours  A  leaves  off,  and  B  finishes  the  work  in  14  hours.  If 
seven  hours  had  occurred  before  A  left  off,  B  would  have  fin- 
ished the  work  in  4§  hours.  How  many  feet  does  each  build 
in  one  hour  ? 

GENERALIZATION   OF  PROBLEMS. 

196.  A  problem  is  said  to  be  generalized  when  letters  are 
used  to  represent  its  known  quantities,  as  well  as  unknown. 

The  unknown  quantities  thus  found  in  terms  of  the  known 
are  general  expressions,  or  formula',  which  may  be  used  for 
the  solution  of  any  similar  problem. 

197.  The  algebraic  solution  of  a  generalized  problem  dis- 
closes many  interesting  truths  and  useful  practical  rules,  as 
may  be  seen  from  the  consideration  of  the  following : 

1.  The  sum  of  two  numbers  is  a,  and  their  difference  is  b ; 
what  are  the  two  numbers  ? 

Let  x  =  the  greater  number, 

and  y  =  the  less. 

By  the  conditions,  x  +  y==a 

x  —  y  =  b 


134  ALGEBRA. 

0  ,  .        ,  .  a  +  b      .  , 

►solving  these  equations,  x  =  — - — ,  the  greater  number, 

i  a  —  b      ,     , 

and  y  =  — - — ,  the  less. 

Hence,  since  a  and  b  may  have  any  value  'whatever,  the 
values  of  x  and  y  are  general,  and  may  be  expressed  as  rules 
for  the  numerical  calculations  in  any  like  case;  thus. 

To  find  two  numbers  when  their  sum  and  difference  are 
given,  —  Add  the  sum  and  difference,  and  divide  by  2,  for  the 
greater  of  tin1  two  numbers;  and  subtract  the  difference  from 
the  sum,  and  divide  by  2,  for  the  less  number. 

For  example,  if  the  sum  of  two  numbers  is  35,  and  their 
difference  13, 

the  greater  = —  =  24, 

and  the  less  =  - — - —  =  11. 

2.  A  can  do  a  piece  of  work  in  a  days,  which  it  requires  b 
days  for  B  to  perform.  In  how  many  days  can  it  be  done  if  A 
and  B  work  together  ? 

Let         x  =  the  number  of  days  required. 

Then       -  =  what  both  together  can  do  in  one  day. 

Also,       -  =  what  A  can  do  in  one  day, 

and         y  =  what  B  can  do  in  one  day. 

By  the  conditions,  -  +  -  =  - 
a       b      x 

Whence,  x  —  -    —  ,  number  of  days  required. 

Hence,  to  find  the  time  for  two  agencies  conjointly  to  ao- 


PROBLEMS.  135 

complish  a  certain  result,  when  the  times  are  given  in  which 
each  separately  can  accomplish  the  same, —  Divide  the  product 
of  the  given  tunes  by  their  sum. 

For  example,  if  A  can  do  a  piece  of  work  in  5  days,  and  B 
in  4  days,  the  time  it  will  take  them  hoth  working  together 

•ill      5x4      20       o,    , 
will  be  -z .  =  —  =  Z%  days. 

5  +  4       9         J      J 

3.  Three  men,  A,  B,  and  C,  enter  into  partnership  for  a 
certain  time.  Of  the  capital  stock,  A  furnishes  m  dollars;  B, 
n  dollars;  and  C, p  dollars.  They  gain  a  dollars.  "What  is 
each  man's  share  of  the  gain  ? 

Let  x  =  A's  share. 

Then,  since  the  shares  arc  proportional  to  the  stocks, 


n  ./• 


in 


=  B's  share, 


and  ==  C's  share. 


m 

ix  x      ty  %c 

By  the  conditions,  x  -\ 1 —  =  a 

in         in 

Whence,  x  = ■ ,  A's  share. 

///  +  n  +  p 

Then,  = ,  B's  share, 

m       m  +  n  +  p       * 

and  — —  = ,  C's  share. 

m        in  +  n  +  p 

Hence,  to  find  each  man's  gain,  when  each  man's  stock  and 
the  whole  gain  are  given,  —  Multiply  the  whole  gain  by  each 
man's  stock,  and  divide  tlisproduct  by  the  whole  stock. 

For  example,  suppose  A's  stock  $300,  B's  $500,  and  C's 
$800,  and  the  whole  gain  $320. 


136 

ALGEBRA. 

Then, 

A's  share 

320  x  300     96000 

=  $60, 

"300  +  500  +  800"   1600 

B's  share 

320  x  500     160000 

=  $100, 

"300  +  500  +  800"  1600 

and  C's  share 

320  x  800    256000 

=  $160. 

300  +  500  +  800        1600 

PROBLEMS. 

4.  A  cistern  can  he  rilled  by  three  pipes  ;  bj  the  first  in  a 
hours,  by  the  second  in  b  hours,  and  by  the  third  in  c  hours. 
In  what  time  can  it  he  filled  by  all  the  pipes  running  together  ? 

5.  Using  the  result  of  the  previous  problem,  suppose  that 
the  first  pipe  fills  the  cistern  in  2  hours,  the  second  in  5  hours, 
and  the  third  in  10  hours.  In  what  time  can  it  be  filled  by 
all  the  pipes  running  together  ? 

6.  Divide  the  number  a  into  two  parts  which  shall  have  to 
each  other  the  ratio  of  m  to  n. 

7.  Using  the  result  of  the  previous  problem,  divide  the 
number  20  into  two  parts  which  shall  have  to  each  other  the 
ratio  of  3  to  2. 

8.  A  courier  left  this  place  n  days  ago,  and  goes  a  miles 
each  day.  He  is  pursued  by  another,  starting  to-day  and 
going  b  miles  daily.  How  many  days  will  the  second  re- 
quire to  overtake  the  first  ? 

9.  In  the  last  example,  if  n  =  3,  a  —  40,  and  b  =  50,  how 
many  days  will  he  required  ? 

10.  Required  what  principal,  at  interest  at  r  per  cent,  will 
amount  to  the  sum  a,  in  t  years  ? 

11.  Using  the  result  of  the  previous  problem,  what  principal, 
at  6  per  cent  interest,  will  amount  to  $3108  in  8  years? 

J.2.    Required  the  number  of  years  in  which  j>  dollars,  at  r 
per  cent  interest,  will  amount  to  a  dollars. 


DISCUSSION   OF   PROBLEMS.  137 

13.  Using  the  result  of  the  previous  problem,  in  how  many 
years  will  $262,  at  7  per  cent  interest,  amount  to  $472.91  ? 

14.  A  banker  has  two  hinds  of  money.  It  takes  a  pieces  of 
the  first  to  make  a  dollar,  and  b  pieces  of  the  second  to  make 
the  same  sum.  If  be  is  offered  a  dollar  for  c  pieces,  how  many 
of  each  kind  must  he  give  ? 

15.  In  the  last  example,  if  a  =  10,  b  =  20,  and  c=  15,  how 
many  of  each  kind  must  he  give  ? 

16.  A  gentleman,  distributing  some  money  among  beggars, 
found  that  in  order  to  give  them  a  cents  each  he  should  want 
b  cents  more;  he  therefore  gave  them  c  cents  each,  and  had  d 
cents  left.     Required  the  number  of  beggars. 

17.  A  mixture  is  made  of  a  pounds  of  coffee  at  m  cents  a 
pound,  h  pounds  at  n  cents,  and  c  pounds  at  j>  cents.  Re- 
quired the  cost  per  pound  of  the  mixture. 

18.  A,  B,  and  C  hire  a  pasture  together  for  a  dollars.  A 
puts  in  m  horses  for  t  months,  B  puts  in  n  horses  for  t'  months, 
and  C  puts  in^j  horses  for  t"  months.  What  part  of  the  ex- 
pense should  each  pay  ? 


XVII.  —  DISCUSSION  OF  PROBLEMS 

LEADING  TO  SIMPLE  EQUATIONS. 

198.  The  Discussion  of  a  problem,  or  of  an  equation,  is  the 
process  of  attributing  any  reasonable  values  and  relations  to 
the  arbitrary  quantities  which  enter  the  equation,  and  inter- 
preting the  results. 

199.  An  Arbitrary  Quantity  is  one  to  which  any  reason- 
able value  may  be  given  at  pleasure. 

200.  A  Determinate  Problem  is  one  in  which  the  given 
conditions  furnish  the  means  of  finding  the  required  quantities. 


138  ALGEBRA. 

A  determinate  problem  leads  to  as  many  independent  equa- 
tions as  tin-re  are  required  quantities  (Art.  195). 

201.  An  Indeterminate  Problem  is  one  in  which  there  are 
fewer  imposed  conditions  than  there  are  required  quantities, 
and,    consequently,    an    insufficient    number    of    independent 

equations  to  determine  definitely  the  values  of  the  required 
quantities. 

202.  An  Impossible  Problem  is  one  in  which  the  condi- 
tions are  incompatible  or  contradictory,  and  consequently  can- 
not be  fulfilled. 

203.  A  determinate  problem,  leading  to  a  simple  equation 
involving  only  one  unknown  quantity,  can  be  satisfied  by  but 
one  value  of  that  unknown  quantity  (Art.  178). 

An  indeterminate  problem,  or  one  leading  to  a  less  number 
of  independent  equations  than  it  has  unknown  quantities,  may 
be  satisfied  by  any  number  of  values. 

For  example,  suppose  a  problem  involving  three  unknown 
quantities  leads  to  only  two  equations,  which,  on  combining, 
give 

x  —  z  =  10, 

or,  x  =  10  +  z. 

Now,  if  we  make  z  =  1,  then  x  =  11 ; 
z  —  2,  then  x  =  12  ; 
z  =  3,  then  x  =  13. 

Thus,  we  may  find  sets  of  values  without  limit  that  will  sat- 
isfy the  equation.     Hence, 

An  indeterminate  equation  may  have  any  manlier  of  so- 
lutions. 

204.  When  a  problem  leads  to  more  independent  equations 
than  it  lias  unknown  quantities,  it  is  impossible. 

For,  suppose  we  have  a  problem  furnishing  three  indepen- 
dent equations,  as, 

x  =  y+l 
y  —  7  —  x 
xy  =  16 


DISCUSSION   OF   PROBLEMS.  139 

From  the  first  two  we  find  x  =  4  and  y  =  3.  But  the  third 
requires  their  product  to  be  10;  hence  the  problem  is  im- 
possible. 

If,  however,  the  third  equation  had  not  been  independent, 
but  derived  from  the  other  two,  as, 

x  y  =  12, 

then  the  problem  would  have  been  possible;  but  the  last  equa- 
tion, not  being  required  for  the  solution,  would  have  been 
redundant. 


INTERPRETATION   OF  NEGATIVE   RESULTS. 

205.  In  a  Negative  Result,  or  a  result  preceded  by  a  — 
sign,  the  negative  sign  is  regarded  as  a  symbol  of  interpre- 
tation. 

Its  significance  when  thus  used  it  is  now  proposed  to  in- 
vestigate. 

1.  Let  it  be  required  to  find  what  number  must  be  added  to 
the  number  a  that  the  sum  may  be  b. 

Let  x  =  the  required  number. 

Then,  a  +  x  =  b 

Whence,  x  =  b  —  a. 

Here,  the  value  of  x  corresponds  with  any  assigned  values  of 
a  and  b.     Thus,  for  example, 

Let  a  =  12,  and  b  =  25. 

Then  x  =  25  -  12  =  13, 

which   satisfies   the   conditions  of  the   problem ;   for  if  13  be 
added  to  12,*  or  a,  the  sum  will  be  25,  or  b. 

But,  suppose       a  =  30,  and  b  =  24. 
Then,  x  =  24  -  30  =  -  0, 


140  ALGEBRA. 

which  indicates  that,  under  the  latter  hypothesis,  the  problem 
is  impossible  in  an  arithmetical  sense,  though  it  is  possible  in 
the  algebraic  sense  of  the  words  "number,"  "added,"  and 
"  sum." 

The  negative  result,  —  6,  points  out,  therefore,  in  the  arith- 
metical sense,  either  an  error  or  "//  impossibility. 

But,  taking  the  value  of  x  with  a  contrary  sign,  we  see  that 
it  will  satisfy  the  enunciation  of  the  problem,  in  an  arithmeti- 
cal sense,  Avhen  modified  so  as  to  read : 

What  number  must  be  taken  from  30,  that  the  remainder 
may  be  24  ? 

2.  Let  it  be  required  to  determine  the  epoch  at  which  A"s 
age  is  twice  as  great  as  B's  ;  A's  age  at  present  being  35  years, 
and  B's  20  years. 

Let  us  suppose  the  required  epoch  to  be  after  the  present 
date. 

Let      x  —  the  number  of  years  after  the  present  date. 

Then,  35  +  x  =  2  (20  +  x) 

Whence,  x  =  —  5,  a  negative  result. 

On  recurring  to  the  problem,  we  find  it  so  worded  as  to 
admit  also  of  the  supposition  that  the  epoch  is  before  the  pres- 
ent date;  and  taking  the  value  of  x  obtained,  with  the  con- 
trary sign,  we  find  it  will  satisfy  that  enunciation. 

Hence,  a  negative  result  here  indicates  that  a  wrong  choice 
was  made  of  two  possible  suppositions  which  the  problem 
allowed. 

From  the  discussion  of  these  problems  we  infer  : 

1.  That  negatire  results  indicate  cither  an  erroneous  enun- 
ciation of  a  'problem,  or  a  wrong  supposition  respecting  the 
quality  of  some  quantity  belonging  to  it. 

2.  That  we  may  form,  when  attainable,  a  'possible  problem 
analogous  to  that   which   involved  the  impassibility,  or  correct 


DISCUSSION   OF   PROBLEMS.  141 

the  wrong  supposition,  by  attributing  to  the  unknown  quan- 
tity in  fin1  equation  a  quality  directly  opposite  to  that 
which  ltml  lice//  attributed  to  it. 

In  general,  it  is  not  necessary  to  form  a  new  equation,  l»ut 
simply  to  change  in  the  old  one  the  sign  of  each  quantity 
which  is  to  have  its  quality  changed. 

Interpret  the  negative  results  obtained,  and  modify  the 
enunciation  accordingly,  in  the  following 


PROBLEMS. 

3.  If  the  length  of  a  field  be  10  rods,  and  the  breadth  8  rods, 
what  quantity  must  be  added  to  its  breadth  so  that  the  con- 
tents may  be  60  square  rods  ? 

4.  If  1  be  added  to  the  numerator  of  a  certain  fraction,  its 
value  becomes  | ;  but  if  1  be  added  to  the  denominator,  it  be- 
comes §.     What  is  the  fraction  ? 

5.  The  sum  of  two  numbers  is  90,  and  their  difference  is 
120  ;  what  are  the  numbers  ? 

6.  A  is  50  years  old,  and  B  40 ;  required  the  time  when  A 
will  be  twice  as  old  as  B- 

7.  A  and  B  were  in  partnership,  and  A  had  3  times  as  much 
capital  as  B.  When  A  had  gained  $  2000.  and  B  $  750,  A  had 
twice  as  much  capital  as  B.  What  was  the  capital  of  each  at 
first  ? 

8.  A  man  worked  14  days,  his  son  being  with  him  6  days. 
and  received  $39,  besides  the  subsistence  of  himself  and  son 
while  at  work.  At  another  time  lie  worked  10  days,  and  had 
his  son  with  him  4  days,  and  received  $28.  What  were  the 
daily  wages  of  each '.' 


142  ALGEBRA. 


XVIII. —  ZERO   AND  INFINITY. 

206.  A  variable  quantity,  or  simply  a  variable,  is  a  quan- 
tity to  which  we  may  give,  in  the  same  discussion,  any  value 
within  certain  limits  determined  by  the  nature  of  the  problem  ; 
a  constant  is  a  quantity  which  remains  unchanged  throughout 
the  same  discussion. 

207.  The  limit  of  a  variable  quantity  is  a  constant  value 
to  which  it  may  be  brought  as  near  as  we  please,  but  which  it 
can  never  reach. 

Thus,  if  3  be  halved,  the  quotient  §  again  halved,  and  so  on 
indefinitely,  the  limit  to  which  the  result  may  be  brought  as 
near  as  we  please,  but  which  it  can  never  reach,  is  zero.  And, 
in  general,  if  any  quantity  be  indefinitely  diminished  by  di- 
vision, its  limiting  value  is  zero. 

208.  If  any  quantity  be  indefinitely  increased  by  multipli- 
cation or  otherwise,  its  limiting  value  is  called  Infinity,  and  is 
denoted  by  the  symbol  co  . 

209.  It  is  evident,  from  the  definition  of  Art.  207,  that  if 
two  variable  quantities  are  always  equal,  their  limiting  values 
will  be  equal. 

210.  We  will  now  show  how  to  interpret  certain  forms 
which  may  be  obtained  in  the  course  of  mathematical  opera- 
tions. 

.  '   a  a 

Let  us  consider  the  fraction  -  :  and  let  -  =  a*. 

b  b 


1.   Interpretation  of 

Let  the  numerator  of        remain   constant,   and   the   denomi- 

h 

nator  be  indefinitely  diminished  by  division.      By    Art.  l-'>7.   it' 
the  denominator  is  divided  by  any  quantity,  the  value  of  the 


ZERO   AND   INFINITY.  143 

fraction  is  multiplied  by  that  quantity ;  hence  the  value  of  the 
fraction,  x,  increases  indefinitely  as  b  is  diminished  indefi- 
nitely. The  limiting  value  of  b  being  0  (Art.  207),  the  limit- 
ing value  of  -   will  be  - ;  and  the  limiting  value  of  x  is  co 

b       a 
(Art.  208).    Now  -  and  x  being  two  variable  quantities  always 

equal,  by  xlrt.  209  their  limiting  values  are  equal ;  or, 


a 


a 
2.   Interpretation  of    — . 

00 


Let  the  numerator  remain  constant,  and  the  denominator  be 
indefinitely  increased  by  multiplication.  By  Art.  lo8,  if  the 
denominator  is  multiplied  by  any  quantity,  the  value  of  the 
fraction  is  divided  by  that  quantity  ;  hence  x  is  diminished 
indefinitely  by  division  as  the  denominator  increases  in- 
definitely.    The    limiting   value    of    b   being  oo,  the  limiting 

ft  ct 

value  of  -   will  be  — :   and  the  limiting  value  of  x  is  0.     By 

b  co 

Art.  209  these  limiting  values  are  equal  ;  or, 

GO 


Problem  of  the  Couriers. 

211.  The  discussion  of  the  following  problem,  commonly 
known  as   that  of   Clairaut,    will    serve    to    further   illustrate 

the  form  -,  besides  furnishing  us  with  an  interpretation  of  the 

,        0 
form  pr . 

Two  couriers,  A  and  B,  are  travelling  along  the  same  road, 
in  the  same  direction,  \V  R,  at  the  rates  of  m  and  n  miles  per 
hour  respectively.     If  at  any  time,  say  12  o'clock,  A  is  at  the 


144  ALGEBKA. 

point  P,  and  B  a  miles  from  him  at  Q,  when  and  where  are 

they  together  '.' 

I ! 1 1 

R'  P  Q  B 

Let  #=  the  required  time  in  hours-, 

and        x  =  the  distance  A  travels  in  the  time  t,  or  the  dis- 
tance from  P  to  the  place  of  meeting. 
Then  x  —  a  =  the  distance  B  travels  in  the  time  t,  or  the  dis- 
tance from  Q  to  the  place  of  meeting. 

Since  the  distance  equals  the  rate  multiplied  by  the  time, 

x  =  m  t 
x  —  a  ==n 1 

Solving  these  equations  with  reference  to  t  and  x, 

a 


x  = 


It  is  proposed  now  to  discuss  these  values  on  different  sup- 
positions. 

1.  in  >  n. 

This  hypothesis  makes  the  denominator  m  —  n  positive; 
hence  the  values  of  both  t  and  x  are  positive.  That  is,  the 
couriers  are  together  after  12  o'clock,  and  to  the  right  of  P. 

This  interpretation  corresponds  with  the  supposition  made. 
For,  if  A  travels  faster  than  B,  he  will  eventually  overtake 
him,  and  in  advance  of  their  positions  at  12  o'clock. 

2.  in  <  n. 

This  hypothesis  makes  the  denominator  m  —  n  negative; 
hence  the  values  of  both  t  and  x  are  negative.  Now.  from 
what  we  have  observed  in  regard  to  negative  results  <  Art.  205), 
these  values  of  t  and  x  indicate  that  the  couriers  w<  re  together 
before  12  o'clock,  and  to  the  left  of  P. 


111  - 

-  n 

m 

a 

m- 

-  n 

ZERO  AND  INFINITY.  145 

This  interpretation  corresponds  with  the  supposition  made. 
For,  if  A  travels  more  slowly  than  B,  he  will  never  overtake 
him ;  but  as  they  are  travelling  along  the  same  road,  they 
must  have  been  together  before  12  o'clock,  and  before  they 
could  have  advanced  as  far  as  P. 

3.    m  =  n. 
This  hypothesis  makes  the  denominator  m  —  n  equal  to  zero  ; 

ft  77h  CL 

so  that  the  values  of  t  and  x  become  -  and  -j—,  respectively; 

or,  by  Art.  210,  t  =  oo  and  x  —  co .  Since  from  its  nature 
(Art.  208),  go  is  a  value  which  we  can  never  reach,  the  values 
of  t  and  x  may  be  regarded  as  indicating  that  the  problem  is 
impossible  under  the  assumed  hypothesis. 

This  interpretation  corresponds  with  the  supposition  made. 
For,  if  the  couriers  were  a  miles  apart  at  12  o'clock,  and  were 
travelling  at  the  same  rate,  they  never  had  been  and  never 
would  be  together. 

Thus,  infinite  results  Indicate  the  imjjossibility  of  a  problem. 

4.    a  =  0,  and  m  >  n  or  m  <  n. 

By  this  hypothesis,  the  values  of  t  and  x  each  become 


m  —  n ' 


or  (Art.  102),  £  =  0  and  x  =  0.  That  is,  the  couriers  are  to- 
gether at  12  o'clock,  at  the  point  P,  and  at  no  other  time  and 
place. 

This  interpretation  corresponds  with  the  supposition  made; 
for,  if  the  distance  between  them  at  12  o'clock  is  nothing,  they 
are  together  at  P ;  but  as  their  rates  are  unequal,  they  cannot 
be  together  after  12  o'clock,  nor  could  they  have  been  together 
before  that  time. 

5.    a  =  0,  and  m  =  n. 

By  this  hypothesis,  the  values  of  t  and  x  each  take  the 

form^. 


146  ALGEBRA. 

Referring  to  the  enunciation  of  the  problem,  we  see  that  if 
the  couriers  were  together  at  12  o'clock,  and  were  travelling 
at  the  same  rate,  they  always  had  been,  and  always  would  be, 
together.  There  is,  then,  no  single  answer,  or  finite  number 
of  answers,  to  the  problem  in  this  case;  and  results  of  this 
form  are  therefore  called  indeterminate. 

Thus,  a  result  -  indicates  indeterminathn. 

212.  The  symbol  -,  however,  does  not  always  represent  an 
indeterminate  quantity  which  may  have  any  fin  ite  mine.  Now, 
in  the  preceding  problem  the  result  -  was  obtained  in  conse- 
quence of  two  independent  suppositions,  one  causing  the  nu- 
merator to  become  zero,  and  the  other  the  denominator.  We 
say  independent^  because  the  quantity  m  —  n  can  be  equal  to  0 
without  necessarily  causing  the  quantity  a  to  become  0.     And 

in  all  similar  cases,  we  should  find  the  result  -  susceptible  of 
the  same  interpretation. 

But  if  the  symbol  -  is  obtained  in  consequence  of  the  same 
supposition  causing  both  numerator  and  denominator  to  be- 
come zero,  it  will  be  found  to  have  a  single  definite  limiting 
value. 

a2  —  b- 
Take,  for  example,  the  fraction  — ;  if  b  =  a,  this  single 

supposition  causes  both  numerator  and  denominator  to  become 
zero,  and  the  fraction  takes  the  form  -. 

Now,  dividing  both  terms  by  a  —  b,  we  have 

o?-V   _a  +  b 

a*-ab~      a      '  {    ' 

which  equation  is  true  so  Ion*;  as  b  is  not  equal  to  a.     It   is 
not  necessarily  true   when  b  is   equal    to  a,  because   the   second 


ZERO   AND   INFINITY.  147 

member  was  obtained  by  dividing  both  terms  of  the  first  mem- 
ber by  a  —  h  (which  divisor  becomes  0  when  b  =  a),  as  we 
cannot  speak  of  dividing  a  quantity  by  nothing. 

In  (1),  as  b  approaches  a,  the  limiting  value  of  the  first 

member  is  --,  and   the   limiting  value   of    the  second   member 

is  2.     Thus  we  have  (Art.  209),  Q  =  2. 

Hence  the  limiting  value  of  the  fraction,  as  b  approaches  a, 
is  2. 

213.     A  proper  understanding  of  the  theory  of  indetermi- 

nation,  and  of  the  relation  of  zero  to  finite  quantities,  will  lead 
to  the  detection  of  the  fallacy  in  some  apparently  remarkable 
results. 

For  example,  let  a  =  b 

Then  ai=  a  b 

Subtracting  //-,  a2  —  J2  =  a  b  —  b'2 

Factoring,  (a  +  b)  (a  —  b)—b  (a  —  b)                        (1) 

Dividing  by  a  —  b,  a  +  b  =  b                                      (2) 

But  b  =  a;  hence  a  +  a  =  a 

then  2  a ■  =  a 

or,  2=1 

The  error  was  made  in  passing  from   (1)  to  (2).     Equation 
(1)  may  be  written 

a  +  b       a  —  b 
b  a  —  b 

Now,  as  b  =  a,  the  second  member  is  an  expression  of  the* form 
- .     But  we  assumed  in  going  from  (1)  to  (2)  that  -     — -  =  1, 

"  '  Cv    "         (J 

or  that  --  =  1 ;  which  we  have  seen  in  Arts.  211  and  212  is  not 
necessarily  the  case,  as  it  may  have  any  value  whatever. 


148  ALGEBRA. 


XIX.  —  INEQUALITIES. 

214.  An  Inequality  is  an  expression  indicating  that  one  of 
two  quantities  is  greater  or  less  than  the  other ;  as, 

a  >  b,  and  m  <  n. 

The  quantity  on  the  left  of  the  sign  is  called  the  first  mem- 
ber, and  that  on  the  right,  the  second  member  of  the  inequality. 

215.  Two  inequalities  are  said  to  subsist  in  the  same  sense 
when  the  first  member  is  the  greater  or  less  in  both. 

Thus, 

a  >  b,  and  c  >  d ;  or  3  <  4,  and  2  <  3, 

are  inequalities  which  subsist  in  the  same  sense. 

216.  Two  inequalities  are  said  to  subsist  in  a  contrary 
sense,  when  the  first  member  is  the  greater  in  the  one,  and 
the  second  in  the  other.     Thus, 

a  >  b,  and  c  <  d ;  or  x  <  y,  and  u  >  z, 

are  inequalities  which  subsist  in  a  contrary  sense. 

217.  In  the  discussion  of  inequalities,  the  terms  greater  and 
less  must  be  taken  as  having  an  algebraic  meaning.     That  is, 

Of"11!/  two  quantities,  a  and  b,  a  is  the  greater  when  a  —  b 
is  positive,  and  a  is  the  less  when  a  —  b  is  negative. 

Hence,  a  negative  quantity  must  be  considered  as  less  than 
nothing;  and,  of  two  negative  quantities,  that  is  the  greater 
which  has  the  least  number  of  units  (Art.  49).     Thus, 

0  >  -  2,  and  -  2  >  -  3. 

218.  An  inequality  will  continue  in  the  same  sense  after 

the  same  quantity  has  been  added  to,  or  subtracted  from,  each 
member. 


INEQUALITIES.  149 

For,  suppose  a  >  b  ; 

then,  by  Art.  217,  a  —  b  is  positive  ;  consequently, 

(a  +  c)  —  (b  +  c)  and  («  —  e)  —  (ft  —  c) 

are  positive,  since  each  equals  a  —  b.     Therefore, 

a  +  c  >  b  +  c,  and  a  —  c  >  S  —  c. 

Hence,  it  follows  that  a  term  may  be  transposed  from  one 
member  of  an  inequality  to  the  other,  if  its  sign  be  changed. 

219.  If  the  signs  of  all  the  terms  of  an  inequality  be 
changed,  the  sign  of  inequality  must  be  reversed. 

For,  to  change  all  the  signs,  is  equivalent  to  transposing 
each  term  of  the  first  member  to  the  second,  and  each  term  of 
the  second  member  to  the  first. 

220.  If  two  or  more  inequalities,  subsisting  in  the  same 
sense,  be  added,  member  to  member,  the  resulting  inequality 
will  also  subsist  in  the  same  sense. 

For,  let 

a>  b,  a'>  V,  a">  b",  


then,  by  Art.  217,  a  —  b,  a'  —  b1,  a"  —  b",  are  all  positive  ; 

and  consequently  their  sum 

a  +  a'  +  a"  + —b  —  b'  —  b"— 

or,  (a  +  a'  +  a"+ )  -  (b  +  V  +  b"  +  ) 

i>  positive.     Hence, 

a  +  a'+  a!'+ >  b  +  b'  +  b"  +  

221.  If  two  inequalities,  subsisting  in  the  same  sense,  be 
subtracted,  member  from  member,  the  resulting  inequality  will 
not  always  subsist  in  the  same  sense. 


150  ALGEBRA. 

For,  let 

a  >  b,  and  a'  >  V ; 

* 

■ 

then  a  —  b  and  a!  —  b1  are  positive  ;  but  a  —  b—  (a'  —  b'),  or 
{a  —  a')  —  (b  —  &')>  may  l)e  either  positive,  negative,  or  0. 

That  is, 

a  —  a'>  b  —  b',  a  —  a'  <  &  —  b',  or  a  —  a'  =  b  —  V. 

222.  -'//  inequality  will  continue  in  the  same  sense  after 
each  member  has  been  multiplied  or  divided  by  the  same  posi- 
tive quantity. 

For,  suppose  a  >  5  ; 

then,  since  «  —  J  is  positive,  if  m  is  positive, 

vi  (a  —  b)  and  —  (a  —  b) 

m  K  ' 

are  positive.     That  is,  m  a—  m  b  and are  positive. 

in       in 

Hence, 

7         i    a         b 
m  a  >  ??i  y,  and  —  >  — . 

in        m 

223.  If  each  member  of  an  inequality  he  multiplied  or  di- 
vided by  the  same  negative  quantity,  the  sign  of  inequality 
must  be  reversed. 

For,  since  multiplying  or  dividing  by  a  negative  quantity 
must  change  the  signs  of  all  the  terms,  the  sign  of  inequality 
must  be  reversed  (Art.  219). 

224.  The  solution  of  an  inequality  consists  in  determining 
the  limit  in  the  value  of  its  unknown  quantity. 

This  may  be  done  by  the  application  of  the  preceding  prin- 
ciples. 

When,  however,  an  inequality  and  an  equation  are  given, 
containing  two  unknown  quantities,  the  process  of  elimination 
will  be  required  in  the  solution. 


INEQUALITIES.  151 

In  verifying  an  inequality,  if  the  symbols  of  the  unknown 
quantities  be  taken  equal  to  their  respective  limits,  the  ine- 
quality becomes  an  equation. 

EXAMPLES. 

225.     1.   Find  the  limit  of  x  in  the  inequality 

23       2x      „ 
i  x  -  -j  >  -g-  +  5. 

Clearing  of  fractions,    21  a;  —  23  >  2  x  +  15 
Transposing,  and  uniting,     19  ic  >  38 
Whence,  x  >  2,  ^4«s. 

2.  Find  the  limits  of  x  in  the  inequalities, 

ax  +  55x-5«i  >a2  (1) 

5x-7ftj;  +  7«Ki'2  (2) 

from  (1),  ax  +  5b  x  >  a2  +  5  ab 

x  (a  +  5  b)  >  a  (a  +  5  b) 
x  >  a. 

From  (2),  bx  —  7ax<b2  —  7ab 

x  (b  -  7  a)  <  b  (b  -  7  a) 
x<  b. 
Hence,  x  is  greater  than  a,  and  less  than  b,  Ant;. 

3.  Find  the  limits  of  x  and  y  in  the  following  inequality  and 
equation  : 

4  x  +  6  y  >  52  (1) 

4  jc  +  2  y  =  32  (2) 

Subtracting  (2)  from  (1),  4  ?/  >  20 

2/  >  5.  (3) 

From  (2),  we  have  y  =  16  —  2  x 


152  ALGEBRA. 

Substituting  in  (3),    16  -  2  x  >  5 

-2x  >-ll 
11 


-*>-- 2 

or   (Art.  219), 

*<T 

Hence, 

y  >  5,  and  # 

4.  Given  5  a?  —  6  >  19.     Find  the  limit  of  cc. 

5.  Given   2x-5  >25;   3x-7<2x+13.       Find  tin- 
limits  of  x. 

6.  Given  3  z  +  1  >  13  —  x;   4cc  —  7  <  2  a;  +  3.     Find  an 
integral  value  of  x. 

7.  Given  5#  +  3?/>46  —  y;  ?/  —  z  =  —  4.     Find  the  lim- 
its of  x  and  y. 

ex  c       d  x  cl 

8.  Given  —-  +  dx  —  crf>T;  -5 c  sc  +  c  d  <  -^-.    Find 

the  limits  of  x. 

9.  Given  2x  +  3y<57;   2x  +  y  =  32.     Find  the  limits 
of  x  and  y. 

10.  A  teacher  being  asked  the  number  of  his  pupils,  replied 
that  twice  their  number  diminished  by  7  was  greater  than  29 ; 
and  that  three  times  their  number  diminished  by  5  was  less 
than  twice  their  number  increased  by  16.  Required  the  num- 
ber of  his  pupils. 

11.  Three  times  a  certain  number,  plus  16,  is  greater  than 
twice  that  number,  plus  24 ;  and  two  fifths  of  the  number,  plus 
5,  is  less  than  11.     Required  the  number. 

12.  A  shepherd  lias  a  number  of  slice])  such  that  three  times 
the  number,  increased  by  2,  exceeds  twice  the  number,  in- 
creased by  61;  and  5  times  the  number,  diminished  by  70,  is 
less  than  4  times  the  number,  diminished  by  9.  How  nianv 
sheep  has  he  ? 


INVOLUTION.  153 


XX.  -  INVOLUTION. 

226.  Involution  is  the  process  of  raising  a  quantity  to  any 
required  power. 

This  may  he  effected,  as  is  evident  from  the  definition  of  a 
power  (Art.  17),  by  taking  the  given  quantity  as  a  factor  as 
many  times  as  there  are  units  in  the  exponent  of  the  required 
power. 

227.  If  the  quantity  to  he  involved  is  positive,  the  signs  of 
all  its  powers  will  evidently  be  positive ;  but  if  the  quantity  is 
negative,  all  its  even  powers  will  be  positive,  and  all  its  odd 
powers  negative.     Thus, 

(—  af  =(-a)x  (—  a)  X  (—  a)  =  +  cr  X  (—  a)  =  —  a3, 

(-  ay  =  (- a)  X  (-  a)  X  (-  a)  x  (-  a)  =  (-  a3)  X  (-  a)  =  +  a\ 

and  so  on. 
Hence, 

Every  even  poiver  is  positive,  and  every  odd  power  has  the 
same  sign  as  its  root. 

INVOLUTION   OF  MONOMIALS. 

228.  1.  Let  it  be  required  to  raise  5  a"  b  c3  to  the  fourth 
power. 

5  a2  bc3x5a2bcsx5aHc3x5a2b  c3  =  625  a8  b*  cv2,  Ans. 

2.    Raise  —3m  n3  to  the  third  power. 
(—  3  m  n3)  X  (—  3  m  n3)  X  (—  3  m  n3)  =  —  21  m3?j9,   Ans. 

RULE. 

Raise  the  numerical  coefficient  to  the  required  power,  and 
multiply  the  exponent  of  each  letter  by  the  exponent  of  the  re- 
(j uired  power  ;  making  the  sign  of  every  wen  poiver  positive, 
and  the  sign  of  every  odd  power  the  same  as  that  of  its  root. 


154  ALGEBRA. 

EXAMPLES. 

j 

Find  the  values  of  the  following  : 

3.  (a2x)2.  7.    (2xm)\  11.    (-2abnxf. 

4.  (-3  cr  b)s.  8.    (2ab2x8)5.  12.    (-7w*ra)4. 

5.  (-ab2c3y.  9.    (a2b2)\  13.    (5a268c4)8. 

6.  (anb)m.  10.    (-a2c3)3.  14.    (-6  a;3?/7)3. 

A  fraction  is  raised  to  any  required  power  by  raising  both 
numerator  and  denominator  to  the  required  power. 
Thus, 

2x*\*_{     2x*\       (     2x2\       f     2x*\  8  a;6 

'3f)  ~\     3y)X\     3y3lX\     3y3)~~     21  if 

Find  the  values  of  the  following : 

15.  m\      it.  L^r.  ia  i  2x^z 


b  J  *  V       3b   I  '  V       36 

16.    R**\\  18.    fL'^V.  20.  f-*'^' 


4  x  y*/  \o         /  V       4  a~ 


INVOLUTION   OF   POLYNOMIALS. 

229.  Polynomials  may  he  raised  to  any  power,  as  is  obvious 
from   Art.   226,  by  the  process  of  successive  multiplications. 

Thus, 

(a  +  b)2=(a  +  b)(a  +  b)  =  a2  +  2ab  +  b'2, 

(a  +  b)3  =(a  +  b)  (a  +  b)  (a  +  b)  =  a3  +  3  a2  b  +  3  a  b2  +  b3, 
and  so  on.     Hence  the  following 

RULE. 

Multiply  the  polynomial  by  itself,  until  it  has  been  taken  as 
a  factor  as  many  times  as  there  are  units  in  the  exponent  of 
the  required  power. 


INVOLUTION.  155 

EXAMPLES. 

Find  the  values  of  the  following : 
1,     (a -b)s.  3.    {l  +  a2  +  b2)\        5.    (am-an)\ 


2.    (|-^)2-  4.    (a  +  m-nf.        6.    (a  +  b) 


5 


In  Chapter  XXXVII  will  be  given  a  method  for  raising  a 
binomial  to  any  required  power,  without  going  through  with 
the  process  of  actual  multiplication. 

SQUARE   OF  A   POLYNOMIAL. 

230.  It  has  been  shown  (Arts.  104  and  105)  that  the 
square  of  any  binomial  expression  can  be  written  down,  with- 
out recourse  to  formal  multiplication,  by  application  of  the 
formulae 

(a  +  b)2  =  a2  +  2ab  +  b2, 

(a-by  =  a?-2ab  +  b2. 
We  may  also  show,  by  actual  multiplication,  that 
(a  +  b  +  c)2=a2  +  2ab  +  2ac  +  b2  +  2bc  +  c2, 

(a  +  b  +  c  +  d)2  =  a2  +  2  ab  +  2  ac  +  2  ad  +  b2  +  2  bc+  2bd 

+  c2  +  2cd  +  d2, 
and  so  on. 

These  residts,  for  convenience  of  enunciation,  may  be  writ- 
ten in  another  form, 

(a  +  b)2  =  a2  +  b2  +  2ab, 

(a  —  b)2  =  a2  +  b2  —  2  a  b, 

(a  +  b  +  c)2  =  a2  +  b2  +  c2  +  2ab  +  2  a  c  +  2b  c, 

(a  +  b  +  c  +  d)2  =  a2  +  b2+c2+d2+2ab  +  2ac  +  2ad 

+  2bc  +  2bd  +  2cd, 

and  so  on.     Hence,  the  following 


156  ALGEBRA. 

RULE. 

Write  the  square  of  each  term,  together  with  twice  its  prod- 
uct by  each  of  the  terms  following  it. 

1.    Square  x2  —  2  x  —  3. 

Square  of  each  term,  a?4  +  4  x2  +9 

Twice  x2  X  the  terms  following,  —  4.x3  —  6x2 

Twice  —  2  x  X  the  term  following,  +  12  x 

» 

Adding,  the  result  is  xA  —  4:X3  —  2x2+12x  +  9. 


EXAMPLES. 

Square  the  following  expressions  : 

2.  a  —  b  +  c.  8.  1  +  x  +  x2  +  x3. 

3.  2x2  +  3x  +  4.  9.  x3-4x2-2x-3. 

4.  2x2-  3x  +  i  10.  2x3+x2+l  x-l. 

5.  a  —  b  —  c  +  d.  l\.  x3  +  bx2  —  x  +  2. 

6.  »i3  +  2a;2  +  a;+2.  12.  3x3-2  x2 -x+ ±. 

7.  1  —  2  a;  +  3  ar.  13.  a  +  &  —  c  —  d  +  e. 

CUBE   OF  A   BINOMIAL. 
231.     Hy  actual  multiplication  we  may  show, 

(a  +  b)3  =  a3  +  3a2b  +  3ab2  +  b3, 
(a-b)3  =  a3-3a2b  +  3a  b2  -  b3. 
Hence,  for  finding  the  cube  of  a  binomial,  the  following 

RULE. 

Write  the  cube  of  the  first  term,  phis  three  times  the  square 
of  the  first  term  times  the  second,  'plus  three  times  the  first 
term  times  the  square  of  the  second,  plus  the  cube  of  the  second 
term. 


3\3 


INVOLUTION.  157 


EXAMPLES. 

1.  Find  the  cube  of  2  x2  -  3  y3. 
(2  a-2)3  +  3  (2  x2)2  (-  3  y3)  +  3  (2  a:2)  (-  3  y3)2  +  (-  3  if) 
'  =  8z6  +  3  (4*4)  (-3t/3)  +  3  (2  a;2)  (9  if)  +  (-27  y9) 

_  8  xG  -  36  x*  f  +  54  x2  if  -  27  y9,  Jns. 

Cube  the  following  : 

2.  a2+2b.  4.   3  a; -4.  6.  4  a;2 -ay. 

3.  2m  +  5«.  5.   2z3-3.  7.   3a;7/  +  5a6 


.2 


CUBE   OF  A   POLYNOMIAL. 
232.     By  actual  multiplication  we  may  show, 

(a  +  ft  +  ey  =  a3  +  b3  +  c3  +  3  a2  b  +  3  a2  c  +  3  b2  a  +  3  b2  c 
+  3  c2  a  +  3c20  +  Gabc, 
(a  +  l)  +  c  +  d)3  =  a3  +  b3  +  c3  +  d3  +  3a2b  +  3a2c  +  3a2d 

+  3b2a  +  3  b2  c  +  3b2  d  +  3  c2  a  +  3  c2  6 
+  3  c2 d  +  3  d2  a  +  3d2b  +  3d2c+6abc 
+  6  a  b  d  +  6  a  c  d  +  6  b  c  d, 
and  so  on.     Hence,  for  finding  the  cube  of  a  polynomial,  the 
following 

BULE. 

Write  the  cube  of  each  term,  tor/ether  with  three  times  the 
product  of  its  square  by  each  of  the  other  terms,  and  also  six 
times  the  product  of  every  three  different  terms. 

EXAMPLES. 

1.    Find  the  cube  of  2  a;2.—  3  x  —  1. 

8  x6  -21  x3  - 1 

-  36  xh  -  12  a4 

-f  54  a;4  -  27  x2 

+    Qx2-9x 
+  36x3 


8  x6  -  36  xs  +  42  x4  +    9  a'3  -  21  x2  -  9  x  -  1,    Arts. 


158  ALGEBRA. 

Find  the  cubes  of  the  following  : 

2.  a  +  b  —  c.  5.  2  -  2  .r  +  x\ 

3.  a;2 -2-1.  6.  1  +  a;  +  x2  +  X3. 

4.  a- 6  +  1.  7.  2.«3-r  +  2x-3. 


XXI.  —  EVOLUTION. 

233.  Evolution  is  the  process  of  extracting  any  required 

root  of  a  quantity. 

This  may  be  effected,  as  is  evident  from  the  definition  of  a 
root  (Art.  17),  by  determining  a  quantity  which,  when  raised 
to  the  proposed  power,  will  produce  the  given  quantity.  It  is, 
therefore,  the  reverse  of  involution. 

234.  Any  quantity  whose  root  can  be  extracted  is  called  a 
perfect  power  ;  and  any  quantity  whose  root  cannot  be  ex- 
tracted is  called  an  imperfect  power. 

A  quantity  may  be  a  perfect  power  of  one  degree,  and  not  of 
another.     Thus,  8  is  a  perfect  cube,  but  not  a  perfect  square. 

235.  To  extract  any  root  of  a  simple  quantity,  the  expo- 
nent of  that  quantity  must  be  divided  by  the  index  of  the  root. 

For,  since  the  ?ith  power  of  am  is  amn  (Art.  228),  it  follows 
that  the  nth.  root  of  amn  is  am. 

236.  Any  root  of  the  product  of  two  or  more  factors  is 
equal  to  the  product  of  the  same  root  of  each  of  the  factors. 

For,  we  have  seen  in  Art.  228,  in  raising  a  quantity  com- 
posed of  factors  to  any  required  power,  that  cadi  of  the  factors 
is  raised  to  the  same  power. 

237.  From  the  relation  of  a  root  to  its  corresponding 
power,  it  follows,  from  Art.  227,  that 


EVOLUTION.  159 

1.  The  odd  roots  of  any  quantity  have  the  same  sign  as  the 
quantity. 

Thus,  \j  a3  =  a  ;  and  ^  —  a5  =  —  a. 

2.  The  even  roots  of  a  'positive  quantity  are  either  positive 
or  negative. 

For  either  a  positive  or  negative  quantity  raised  to  an  even 
power  is  positive.     Thus, 

y/  a4  =  a  or  —  a  ;  or,  y'  a*  =  ±  a. 

Note.  The  sign  ±,  called  the  double  sign,  is  prefixed  to  a  quantity 
■when  we  wish  to  indicate  that  it  is  either  +  or  -  . 

3.  Even  roots  of  a  negative  quantity  are  not  possible. 

For  no  quantity  raised  to  an  even  power  can  produce  a  neg- 
ative result.  Such  roots  are  called  impossible  or  imaginary 
quantities. 

EVOLUTION   OF   MONOMIALS. 

238.  From  the  principles  contained  in  Arts.  235  to  237, 
we  obtain  the  following 

RULE. 

Extract  the  required  root  of  the  numerical  coefficient,  ami 
divide  the  exponent  of  each  letter  by  the  index  of  the  root; 
making  the  sign  of  every  even  root  of  a  positive  quantity  ±, 
and  the  sign  of  every  odd  root  of  any  quantity  the  same  as 
that  of  the  quantity. 

If  the  given  quantity  is  a  fraction,  it  follows  from  Art.  228 
that  we  may  fake  the  required  root  of  both  of  its  terms. 

EXAMPLES. 

1.    Find  the  square  root  of  9  a4  b-  c6. 


\/9a4i2c6  =  ±3fl2i  cs,  Ans. 


2.    Find  the  cube  root  of  -  64  a9  xz  y6. 

.3/ 


^  _  64  a9  xs  yG  =  —  4:a3x  y2,  Ans. 


160  ALGEBRA. 


8  X3  7)1 

3.    Find  the  cube  root  of 


27  aG  b9 


8  /  /8  x3  m12\  _  2  x  m4 


Find  the  values  of  the  following : 

* 

4.  if  -  125  x3  y6.         9.  yV"1^.  14.  \/  729  a1*  b24  <A 


5.  ySla*b8.  10.  Sl-8an«x\       15.  \/-32  c6"^0™. 

_     J  f  32  m5n10\  . 5/ 

V  \     243     J  '      1L  ^ 16  a;2m+2  a2"     16-  V  243  w16  »*■ 

7.  \/l»^?.  12.  y/^QQ^l)  •    17.  V(«  +  *)W- 

8.  y/  625  a12  c2.  13.  y'  32"  63n  an.  18.  faj8B  +  V1"6- 

SQUARE   ROOT   OF   POLYNOMIALS. 

239.  In  Art.  11G  we  explained  a  method  of  extracting  the 
square  root  of  a  trinomial,  provided  it  was  a  perfect  square. 
We  will  now  give  a  method  of  extracting  the  square  root  of 
any  polynomial  which  is  an  exact  square. 

Since  the  square  of  a  +  b  is  a2  +  2  a  b  +  b2,  we  know  that 

the  square  root  of  a2  +  2  a  b  +  b2  is  a  +  b.     If  we  can  discover 

an  operation  by  which  we  can  derive  a  +  b  from  a2  +  2  a  b  +  b2, 

we  can  give  a  rule  for  the  extraction  of  the  square  root. 

„       „      7       7„  7  Arranmne   the    terms    of    the 

a2  +  2ab+b2   a  +  b  5    °.  ' 

2  square  according  to  the  descend- 

o  „  ,   7.     2  a  b  A-  V1  *n8  powers  of  a,  we  observe  thai 

2  a  b  +  b2  the  square  root  of  the  first  term, 

a2,  is  a,  which  is  the  first  term 
of  the  required  root.  Subtract  its  square,  a2,  from  the  uiven 
polynomial,  and  bring  down  the  remainder,  2  a  b  +  b2  or 
(2  a  +  b)  b.  Dividing  the  first  term  of  the  remainder  by  2  a, 
that  is,  by  twice  the  first  term  of  the  root,  we  obtain  b,  the 
other  term.     This,  added  to  2  a,  completes  the  divisor,  2  a  +  b  ; 


EVOLUTION.  161 

which,  multiplied  by  b,  and  the  product,  2  ab  +  V2,  subtracted 
from  the  remainder,  completes  the  operation. 

By  a  similar  process,  a  root  consisting  of  more  than  two 
terms  may  be  found  from  its  square.  Thus,  by  Art.  230,  we 
know  that  (a  +  b  +  c)2  =  a2  +  2  a  b  +  b2  +  2  a  c  +  2  b  c  +  c2. 
Hence,  the  square  root  of  a2  +  2  a  b  +  b'2  +  2  a  c  +  2  b  c  +  c2  is 
a  +  b  +  c. 

a2  +  2ab  +  b2  +  2ac+2bc  +  c2 

•2 

a 


a  +  b  +  c 


2a  +  b 


2ab  +  b2  +  2ac+2bc  +  c2 
2ab  +  b2 


2  a  +  2  b  +  c 


2ac+2bc  +  c2 
2ac+2be  +  r2 


The  square  root  of  the  first  term,  a2,  is  a,  which  is  the  first 
term  of  the  required  root.  Subtracting  a2  from  the  given  poly- 
nomial, we  obtain  2  a  b  as  the  first  term  of  the  remainder. 
Dividing  this  by  twice  the  first  term  of  the  root,  2  a,  we  ob- 
tain the  second  term  of  the  root,  b,  which,  added  to  2  a,  com- 
pletes the  divisor,  2  a  +  b.  Multiplying  this  divisor  by  b,  and 
subtracting  the  product,  2  a  b  +  b'2,  from  the  first  remainder, 
we  obtain  2  a  c  as  the  first  term  of  the  next  remainder. 

Doubling  the  root  already  found,  giving  2  a  +  2  b,  and  di- 
viding the  first  term  of  the  second  remainder,  2  a  c,  by  the  first 
term  of  the  result,  2  a,  we  obtain  the  last  term  of  the  root,  c. 
This,  added  to  2  a  +  2  b,  completes  the  divisor,  2  a  +  2  b  +  c ; 
which,  multiplied  by  the  last  term  of  the  root,  c,  and  subtracted 
from  the  second  remainder,  leaves  no  remainder. 

From  these  operations  we  derive  the  following 

RULE. 

Arrange  the  terms  according  to  the  powers  of  some  letter. 

Find  the  square  root  of  the  first  term,  write  it  as  the  first 
term  of  the  root,  and  subtract  its  square  from  the  given  poly- 
nomial. 

Divide  the  first  term  of  the  remainder  by  double  the  root 
already  found,  and  add  the  result  to  the  root,  and  also  to  the 
divisor. 


162 


ALGEBRA. 


Multiply  the  divisor  as  it  now  stands  by  the  term  of  the  root 
last  obtained,  and  subtract  the  product  from  the  remainder. 

If  there  are  other  terms  remaining,  continue  the  operation 
In  the  same  manner  as  before. 

Note.  Since  all  even  roots  have  the  double  sign  ±  (Art.  237),  all  the 
terms  of  the  result  may  have  their  signs  changed.  In  the  examples,  how- 
ever, we  shall  consider  only  the  positive  sign  of  the  result. 


EXAMPLES. 

1.    Find  the  square  root  of  9  x*  —  12  xs  +  16  x2  —  8  x  +  4. 

9a;4-12a;3  +  lGx~-8x  +  4:    3a;2-2a;  +  2 

9  a;4 


6  a:2  -  2 


./' 


-12  a;3 
- 12  x3 


4  a;'2 


6  x-  -  4  x  +  2 


12  x2  -  8  x  +  4 
12  a;'2-.  8  x  +  4 


Ans.  3x2  —  2x  +  2. 
Find  the  square  roots  of  the  following : 


2.    ±xi-±xs-3x2  +  2x  +  l. 


2        1 

—  +  — i 
m       m 


3.  4  a4 -16  a3 +24  a2- 16  a +  4. 

4.  m2  +  2  m  —  1 

5.  9  — 12  x  +  10  :•■-  -  4  x3  +  x*. 

6.  19  x2  +  6  x3  +  25  +  xi  +  30  a-. 

7.  28  a-3  +  4  a;4  -  14  x  +  1  +  45  x2. 

8.  40  jc  +  25  - 14  x2  +  9  x*  -  24  a;3. 

9.  4  .r4  +  64  -  20  a-3  -  80  x  +  57  x2. 

10.  a2  +  b2  +  c2 -2  ab  -2  a  c  +  2  b  c 

11.  a;2  +  4  y2  +  9  «2  —  4  a;  ?/  +  6  a;  »  —  12  y  z. 

No  rational  binomial  is  an  exact  square;;  hut,  by  the  rule, 
the  (ip/imxliiinte  root  may  be  found. 


EVOLUTION.  10- 


•  i 


Find,  to  four  terms,  the  approximate  square  roots  of  the  fol- 
lowing : 
12.   1  +  x.  13.    a2  +  b.  14.    1  —  2  x.  15.    a2  +  x2. 

The  square  root  of  a  perfect  trinomial  square  may  be  ob- 
tained by  the  rule  of  Art.  116, 

Find  the  square  roots  of  the  first  and  last  terms,  and  con- 
nect the  results  by  the  sign  of  the  second  term. 

Extract  the  square  roots  of  the  following : 

16.  x4  +  8x2+16.  19.    «2m  -4  am+n  +  4  a2". 

rtrt     a2       4  a      ±b2 

17.  9x*-6xf  +  f.  20.   _-  — +^. 

t2  4  x2  9  ?/4 

18.  a*-ax  +  T.  8L9?  +  2+4^- 

SQUARE   ROOT   OF   NUMBERS. 

240.  The  method  of  Art.  239  may  be  used  to  extract  the 
square  roots  of  numbers. 

The  square  root  of  100  is  10  ;  of  10000  is  100  ;  of  1000000, 
is  1000 ;  and  so  on.  Hence,  the  square  root  of  a  number  less 
than  100  is  less  than  10  ;  the  square  root  of  a  number  between 
10000  and  100  is  between  100  and  10 ;  the  square  root  of  a 
number  between  1000000  and  10000  is  between  1000  and  100  ; 
and  so  on. 

Or,  in  other  words,  the  integral  part  of  the  square  root  of  a 
number  of  one  or  two  figures,  contains  one  figure;  of  a  number 
of  three  or  four  figures,  contains  two  figures ;  of  a  number  of 
five  or  six  figures,  contains  three  figures ;  and  so  on.     Hence, 

If  a  point  is  placed  over  evei*y  second  figure  in  any  integral 
number,  beginning  with  the  units'1  place,  the  number  of  point* 
will  shoiv  the  number  of  figures  in  the  integral  part  of  its 
square  root. 


164  ALGEBRA. 


241.     Let  it  be  required  to  find  the  square  root  of  4356. 

Pointing  the  number  according  to 

60+6     the  rule  of  Art.  240,  it  appears  that 

there  are  two  figures  in  the  integral 


4356 
3600 


120  +  6    756  'J"  °  ~~— -~0. 

jgQ  part  oi  the  square  root.     Let  a  denote 

the  figure  in  the  tens'  place  in  the 
root,  and  b  that  in  the  units'  place.  Then  a  must  be  the 
greatest  multiple  of  10  whose  square  is  less  than  4356 ;  this 
we  find  to  be  60.  Subtracting  a2,  that  is,  the  square  of  60,  or 
3600,  from  the  given  number,  we  have  the  remainder  756. 
Dividing  this  remainder  by  2  a,  or  120,  gives  6,  which  is  the 
value  of  b.  Adding  this  to  120,  multiplying  the  result  by  6, 
and  subtracting  the  product,  756,  there  is  no  remainder. 
Therefore  we  conclude  that  60  +  6,  or  66,  is  the  required  square 
root. 

The  zeros  being  omitted  for  the  sake  of  brevity,  we  may  ar- 
range the  work  in  the  following  form  : 


4356 
36 


G6 


126 


756 
756 


RULE. 

Separate  the  given  number  into  periods,  by  pointing  every 
second  figure,  beginning  with  the  units'  [dace. 

Find  the  greatest  square  in  the  left-hand  period,  and  place 
its  root  on  the  right  ;  subtract  the  square  of  this  root  from  the 
first  period,  and  to  the  remainder  bring  down  the  next  period 
for  a  dlr  hi  end. 

Divide  this  dividend,  omitting  the  last  figure,  by  double  the 
root  already  found,  and  annex  the  result  to  the  root  and  also 
to  the  divisor, 

Multiply  tin'  divisorj  as  it  now  stands,  by  the  figure  of  the 
root  last  obtained,  and  subtract  the  product  from  tin-  dividend. 

If  there  are  more  periods  to  be  brought  down,  continue  the 
operation  in  the  same  manner  as  before. 


EVOLUTION.  165 

If  there  be  a  final  remainder,  the  given  number  has  not  an 
exact  square  root ;  and,  since  the  rule  applies  equally  to  deci- 
mals, we  may  continue  the  operation,  by  annexing  periods  of 
zeros  to  the  given  number,  and  thus  obtain  a  decimal  part  to 
be  added  to  the  integral  part  already  found. 

It  will  be  observed  that  decimals  require  to  be  pointed  to 
the  right ;  and  if  they  have  no  exact  root,  we  may  continue 
to  form  periods  of  zeros,  and  obtain  decimal  figures  in  the  root 
to  any  desirable  extent. 

As  the  trial  divisor  is  necessarily  an  incomplete  divisor,  it  is 
sometimes  found  that  after  completion  it  gives  a  product' larger 
than  the  dividend.  In  such  a  case,  the  last  root  figure  is  too 
large,  and  one  less  must  be  substituted  for  it. 

The  root  of  a  common  fraction  may  be  obtained,  as  in  Art. 
238,  by  taking  the  root  of  both  numerator  and  denominator, 
when  they  are  perfect  squares.  If  the  denominator  only  is  a 
perfect  square,  take  the  approximate  square  root  of  the  nu- 
merator, and  divide  it  by  the  square  root  of  the  denominator. 
If  the  denominator  is  not  a  perfect  square,  either  reduce  the 
fraction  to  an  equivalent  fraction  whose  denominator  is  a  per- 
fect square,  or  reduce  the  fraction  to  a  decimal. 

EXAMPLES. 

1.    Extract  the  square  root  of  49.434961. 


49.434961 
49 


7.031 


1403 


4349 
4209 


14061 


14061 
14061 


Ans.  7.031. 

Here  it  will  be  observed  that,  in  consequence  of  the  zero  in 
the  root,  we  annex  one  zero  to  the  trial  divisor,  14,  and  bring 
down  to  the  corresponding  dividend  another  period. 

Extract  the  square  roots  of  the  following : 


ALGEBRA. 

6. 

.9409. 

10. 

.006889. 

7. 

6561 
9025' 

11. 

.0000107584 

8. 

1.170724. 

12. 

811440.64. 

9. 

446.0544. 

13. 

.17015625. 

166 

2.  273529. 

3.  45796. 

4.  106929. 

5.  33.1776. 

Extract  the  square  roots  of  the  following  to  the  fifth  decimal 
place : 

14.  2.  16.   31.  18. 

15.  5.  17.   173.  19. 

242.  When  n  +  1  figures  of  a  square  root  have  been  ob- 
tained by  the  ordinary  method,  n  more  may  be  obtained  by 
simple  division  only,  supposing  2n  +  l  to  be  the  whole  num- 
ber. 

Let  N  represent  the  number  whose  square  root  is  required, 
a  the  part  of  the  root  already  obtained,  x  the  rest  of  the  root ; 
then 

y/_ZV=  a  +  x, 
whence,  iV=  a2  +  2  a  x  +  x2 ; 

therefore,  JV —  a2  =  2  a  x  +  x2, 


7 

1 

20. 

9' 

3' 

3 

2 

21. 

16' 

N-a* 


=  x  + 


x~ 


2  a  '2a 

Then  iV—  a2  divided  by  2  a  will  give  the  rest  of  the  square 

x 
root  required,  or  x,  increased  by  -= — ;  and  we  shall  show  that 

x 
-= —  is  a  proper  fraction,  less  than  \,  so  that  by  neglecting  the 

Zi  a 

remainder  arising  from  the  division,  we  obtain  the  part  re- 
quired. For,  x  by  supposition  contains  n  figures,  so  that  x2 
cannot  contain  more  than  2  n  figures  ;  but  a  contains  2n  +  l 


EVOLUTION. 


167 


figures ;    and  hence   —    is   a  proper  fraction.     Therefore  - 

a  2  a 

is  a  proper  fraction,  and  less  than  ^. 

In  the  demonstration  we  supposed  JV  an  integer  with  an 
exact  square  root ;  but  the  result  may  be  extended  to  other 
cases. 

From  the  examples  in  Art.  241,  we  observe  that  each  re- 
mainder brought  down  is  the  given  expression  minus  the 
square  of  the  root  already  obtained ;  and  is  therefore  in  the 
form  A7" — a2.  If,  then,  any  remainder  be  divided  by  twice 
the  root  already  found,  we  can  obtain  by  the  division  as  many 
more  figures  of  the  root  as  we  already  have,  less  one. 

We  will  apply  these  principles  to  calculating  the  square  root 
of  12  to  the.  sixth  decimal  place.  We  will  obtain  the  first  four 
figures  of  the  result  by  the  ordinary  method : 


12.000000 
9 


3.464 


64 


300 
256 


686 


4400 
4116 


6024 


28400 

27696 


'04 


The  remainder  now  is  .000704 ;  and  twice  the  root  already 
found  is  6.928.  Then,  by  dividing  .000704  by  6.928,  we  can 
obtain  the  next  three  figures  of  the  root.     Thus, 

6.928). 0007040  (.000102 
.0006928 

11200 

That  is,  the  square  root  of  12  to  the  nearest  sixth  decimal 
place  is  3.464102. 

The  following  rule  will  be  found  to  save  trouble  in  obtaining 
approximate  square  roots  by  this  method : 


168 


ALGEBRA. 


Divide  the  remainder  by  twice  the  root  already  found  {omit- 
ting the  decimal  point),  and  annex  all  of  the  quotient,  except 
the  decimal  point,  to  the  part  of  the  root  already  found. 

In  practice  the  work  would  be  arranged  thus : 


12. 
9 

64  30C 

25C 

3.464 

) 

686 

4400 
4116 

6924 

:  28400 
27696 

6928)  704.000  (.102 
6928 

11200 

Am.  3.464102 

EXAMPLES. 

1.  Extract  the  square  root  of  11  to  the  4th  decimal  place. 

2.  Extract  the  square  root  of  3  to  the  6th  decimal  place. 

3.  Extract  the  square  root  of  61  to  the  8th  decimal  place. 

4.  Extract  the  square  root  of  131  to  the  3d  decimal  place. 

5.  Extract  the  square  root  of  781  to  the  5th  decimal  place. 

6.  Extract  the  square  root  of  12933  to  the  4th  decimal  place. 


CUBE   ROOT   OF   POLYNOMIALS. 

243.     Since  (a  +  b)3  =  as  +  3  a2  b  +  3  a  b2  +  b3,  we  know 
that  the  cube  root  of  a3  +  3  a"  b  +  3  a  b2  +  b3  is  a  +  b. 


a3  +  3  a2  b  +  3  a  b2  +  b3 

a" 


a  +  b 


3a2+3ab  +  b2    3  a2  b  +  3  a  b2  +  b* 
3a2b  +  3ab2  +  b5 


EVOLUTION.  1G9 

Arranging  the  terms  of  the  cube  according  to  the  descending 
powers  of  a,  we  observe  that  the  cube  root  of  the  first  term,  a3, 
is  a,  which  is  the  first  term  of  the  required  root.  Subtract  its 
cube,  a3,  from  the  given  polynomial,  and  bring  down  the  re- 
mainder, 3  a2  b  +  3  a  b'2  +  b3  or  (3  a2  +  3  a  b  +  b'2)  b.  Dividing 
the  first  term  of  the  remainder  by  3  a2,  that  is,  by  three  times 
the  square  of  the  first  term  of  the  root,  we  obtain  b,  the  other 
term  of  the  root.  Adding  to  the  trial  divisor  3  a  b,  that  is, 
three  times  the  product  of  the  first  term  of  the  root  by  the  last, 
and  b'2,  that  is,  the  square  of  the  last  term  of  the  root,  completes 
the  divisor,  3  a'2  +  3  a  b  +  b'2 ;  which,  multiplied  by  b,  and  the 
product,  3  a2  b  +  3  a  b'2  +  b3,  subtracted  from  the  remainder, 
completes  the  operation. 

If  there  were  more  terms,  we  should  proceed  with  a  +  b 
exactly  as  previously  with  a ;  regarding  it  as  one  term,  and 
dividing  the  first  term  of  the  remainder  by  three  times  its 
square  ;  and  so  on.     Hence,  the  following 

RULE. 

Arrange  the  terms  according  to  the  powers  of  some  letter. 
Find  the  cube  root  of  the  first  term,  write  It  as  the  first  term 
of  the  root,  and  subtract  its  cube  from  the  given  polynomial. 

Take  three  times  the  square  of  the  root  already  found  for  a 
trial  divisor,  divide  the  first  term  of  the  remainder  by  it,  and 
write  the  quotient  for  the  next  term  of  the  root. 

Add  to  the  trial  divisor  three  times  the  prod  net  of  the  first 
term  by  the  second,  and  the  square  of  the  second  term. 

Multiply  the  complete  divisor  by  the  second  term  of  the  root, 
and  subtract  the  product  from  the  remainder. 

If  there  are  other  terms  remaining,  consider  the  root  already 
found  as  one  term,  and  proceed  as  before. 


EXAMPLES. 

1.    Find  the  cube  root  of   x6  —  6  x'°  +  40  x3  —  96  x  —  64. 


170 

X 

X 

ALGEBRA. 
6-6a5+40a;3-96a-64 1  x2-2j 

6 

3xi-6xa+4:xi 

—6  a-5 

— 6  a;5  +12  a4-  8  xs 

3  a4  - 12  a3  +  12  a;2 

-12a2+24a;  +  16 

-12a;4+48a;3 

3a4-12a:3             +  24a 

+  16 

- 12  x* +48  x3-  96  .r  -64 

Ans.  x'2  —  2  a;  —  4. 

The  formation  of  the  second  divisor  may  be  explained  thus : 

Regarding  the  root  already  obtained,  a2  —  2  a,  as  one  term, 
three  times  its  square  gives  3  a4  —  12  a3  +  12  x'2 ;  three  times 
x~  —  2  a;  times  —  4;  gives  —  12  x'2  +  24  x  ;  and  the  square  of  the 
last  root  term  is  16.  Adding  these  results,  we  have  for  the 
complete  divisor,  3  a4  —  12  x3  +  24  x  +  16. 

Find  the  cube  roots  of  the  following: 

2.  1  —  6  y  +  12  y'2  -  8  ?/3. 

3.  8  a;6  +  36  a;4  +  54  a;2  +  27. 

4.  64  a;3  -  144  a  b  x2  +  108  a2  i2  a:  -  27  a3  63. 

5.  a-6  +  6  a-5  -  40  a3  +  96  x  —  64. 

6.  v/«-l  +  57/3-37/5-3y. 

7.  a:3  +  3a;H h-g. 

8.  15  r4  -  6  a-  -  6  a;5  +  15  x2  +  1  +  a6  -  20  x*. 

9.  aa  +  3  aa  &  +  3a2c  +  3aJ2  +  6a6c  +  3ac2+S8  +  3  6a  c 

+  3l>  r2  +  c8. 

10.  9  a3  -  21  x2  -  36  a5  +  8  a;6  -  9  x  +  42  a-4  -  1. 

No  rational  binomial  is  an  exact  cube;  but,  by  the  rule, 
the  approximate  root  may  be  found. 


EVOLUTION. 


171 


Find,   to  four  terms,   the  approximate    cube  roots   of   the 
following : 


11.    Xs +  1. 


12.   xs  — 


a° 


13.    8 


./• 


o 


CUBE   ROOT  OF  NUMBERS. 

244.  The  method  of  Art.  243  may  he  used  to  extract  the 
cube  roots  of  numbers. 

The  cube  root  of  1000  is  10;  of  1000000,  is  100;  of 
1000000000,  is  1000;  and  so  on.  Hence,  the  cube  root  of  a 
number  less  than  1000  is  less  than  10  ;  the  cube  root  of  a  num- 


ber between  1000000  and  1000  is  between  100  and  10 ;  the 
cube,  root  of  a  number  between  1000000000  and  1000000  is 
betwTeen  1000  and  100 ;  and  so  on. 

Or,  in  other  words,  the  integral  part  of  the  cube  root  of  a 
number  of  one,  two,  or  three  figures,  contains  one  figure;  of 
a  number  of  four,  five,  or  six  figures,  contains  two  figures ; 
of  a  number  of  seven,  eight,  or  nine  figures,  contains  three 
figures ;  and  so  on.     Hence, 

If  a  point  is  placed  over  every  third  figure  in  any  integral 
a  miller,  beginning  with  the  units'  place,  the  number  of  points 
will  show  the  number  of  figures  lit  the  integral  part  of  its  eube 
root. 

245.     Let  it  be  required  to  find  the  cube  root  of  405224. 

Pointing  the  number  according  to 
the  rule  of  Art.  244.  it  appears  that 
there  are  two  figures  in  the  integral 
part  of  the  cube  root.  Let  '/  denote 
the  figure  in  the  tens'  place  in  the 
root,  and  b  that  in  the  units'  place. 


405224 
343000 


70  +  4 


14700 

840 

16 


15556 


62224 


62224 


Then   a   must   be  the    greatest   mul- 


tiple of  10  whose  cube  is  less  than  405224 ;  this  we  find  to 
be  70.  Subtracting  a3,  that  is,  the  cube  of  70,  or  343000, 
from  the  given  number,  we  have  the  remainder  62224.  Divid- 
ing this  remainder  by  3  a'2,  or  14700,  gives  4,  which  is  the 


172  ALGEBRA- 

value  of  b.  Adding  to  the  trial  divisor  3  a  b,  which  is  840, 
and  b'2,  which  is  16,  completes  the  divisor,  15556.  Multiplying 
the  result  by  4,  and  subtracting  the  product,  02224,  there  is 
no  remainder.  Therefore  we  conclude  that  70  +  4,  or  74,  is 
the  required  cube  root. 

The  work  is  usually  arranged  thus : 


405224 
343 


74 


14700 

840 

16 

15556 


62224 


62224 


RULE. 

Separate  the  given  member  into  periods,  by  pointing  every 
third  figure,  beginning  ivith  the  units'  place. 

Find  the  greatest  cube  in  the  left-hand  period,  and  place  its 
root  on  the  right /  subtract  the  cube  of  this  root  from  the  left- 
hand  period,  and  to  the  remainder  bring  down  the  next  period 
for  a  dividend. 

Divide  this  dividend,  omitting  the  last  two  figures,  by  three 
times  the  square  of  the  root  already  found,  and  annex  the  quo- 
tient to  the  root. 

Add  together  the  trial  divisor,  with  two  zeros  annexed; 
three  times  the  product  of  the  last  root  figure  by  the  rest  of  the 
root,  ivith  one  zero  annexed ;  and  the  square  of  the  last  root 
figure. 

Multiply  the  divisor,  as  it  now  stands,  by  the  figure  of  the 
root  last  obtained,  and  subtract  the  product  from  the  dividend. 

If  there  are  more  periods  to  be  brought  down,  continue  the 
operation  in  the  same  manner  as  before,  regarding  the  root 
already  obtained  as  one  term. 

The  observations  made  after  the  rule  for  the  extraction  of 
the  square  root  (Art.  241)  are  equally  applicable  to  the  extrac- 
tion of  the  cube  root. 


EVOLUTION. 


173 


EXAMPLES. 

1    Extract  the  cube  root  of  8.144865728. 


8.144865728 
8 

2.012 

120000 

600 

1 

120601 

144865 
120601 

12120300 

12060 

4 

24264728 
24264728 

12132364 

Ans.  2.012. 

Here  it  will  be  observed  that,  in  consequence  of  tbe  0  in 
the  root,  we  annex  two  additional  zeros  to  the  trial  divisor, 
1200,  and  bring  down  to  tbe  corresponding  dividend  another 
period. 


Extract  the  cube  roots  of  the  following : 


2.   1860867. 


4.   1481.544. 


6.   51.478848. 


3.   .724150792. 


29791 
681472 


7.   .000517781627. 


Extract  the  cube  roots  of  the  following  to  the  third  decimal 


place : 


8.  3. 

10.  212. 

124 

9.  7. 

11.  5 

8 

13  3 
13'  17 

246.  JVJien  n  +  2  figures  of  a  cube  root  have  been  obtained 
by  the  ordinary  method,  n  more  may  be  obtained  by  division 
only,  supposing  2  n  +  2  to  be  the  whole  number. 


174  ALGEBRA. 

Let  N  represent  the  number  whose  cube  root  is  required,  a 
the  part  of  the  root  already  obtained,  x  the  rest  of  the  root ; 
then 

$  2V=  a  +  x, 

whence,  iV=  a3  +  3  a2  x  +  3  a  x2  +  xs ; 

therefore,  _Ar—  a3  =  3  a2  x  +  3  a  x2  +  x3, 

iV" —  a3  x2       x3 


»  +  — + 


dec  a       6  a- 

Then  A7"—  a3  divided  by  3  a2  will  give  the  rest  of  the  cube 


root  required,  or  x,  increased  by  - 1-  77—,;  and  we  shall  show 

a       o  a 

that  the  latter  is  a  proper  fraction,  less  than  h,  so  that  by 
neglecting  the  remainder  arising  from  the  division,  we  obtain 
the  part  required.  For,  x  by  supposition  contains  n  figures, 
so  that  x2  cannot  contain  more  than  2  n  figures.     But  «  con- 

o 

X" 

tains  2  n  +  2  figures  ;  and  hence  - —  is  less  than  ^  .     And  as 

n — 5=  —  X  o — ,  and  - —  is  less  than  1,  - — .  must  also  be  less 
o  cr       a        6  a  6  a  -6  a1 

than  TTn.     Therefore, (-  ^ — ^  is  a  proper  fraction,  less  than  \. 

a       k>  a 

Remarks  similar  to  those  in  the  last  part  of  Art.  242  apply 
here. 


ANY  ROOT   OF   POLYNOMIALS. 

247.  In  order  to  establish  a  general  rule  for  the  extraction 
of  roots,  it  will  be  necessary  to  notice  the  formation  of  the  n\\\ 
power  of  a  polynomial,  n  being  any  entire  number  whatever. 
Thus, 

(a  +  by  =  an  +  n  an~}  b  + 

Therefore, 

y'  an  +  n  au~l  b  + =  a  +  b. 

The  first  term  of  the  root,  a,  is  the  nth  root  of  an,  the  first 
term  of  the  power;  and  the  .second  term  of  the  root,  b,  may  be 


EVOLUTION.  175 

obtained  by  dividing  the  second  term  of  the  power  by  n  an~\ 
or  by  n  times  the  (n  —  l)th  power  of  the  first  term  of  the  root. 

If  the  root  now  found  be  raised  to  the  nth  power,  and  sub- 
tracted from  the  given  polynomial,  it  will  be  seen  that  two 
terms  of  the  required  root  have  been  determined. 

It  will  be  observed  that  the  process  is  essentially  that  of  the 
preceding  Articles,  simplified  by  dispensing  with  completed 
divisors,  and  generalized.     Hence  the  • 

RULE. 

Arrange  the  terms  according  to  the  powers  of  some  letter. 

Find  the  required  root  of  the  first  term  for  the  first  term  of 
the  root,  and  subtract  its  power  from  the  given  polynomial. 

Divide  the  first  term  of  the  remainder  by  n  times  the 
(n  —  l)th power  of  this  root,  for  the  second  term  of  the  root, 
and  subtract  the  nth  power  of  the  root  now  found  from  the 
given  polynomial. 

If  other  terms  of  the  root  require  to  be  determined,  use  the 
same  divisor  as  before,  and  proceed  in  like  manner  till  the  nth 
poiver  of  the  root  becomes  equal  to  the  given  polynomial. 

This  rule  is,  also,  applicable  to  numbers,  by  taking  n  figures 
in  each  period. 

EXAMPLES. 

1.   Extract  tl*  cube  root  of  x6  +  6  x5  —  40  xs  +  96  x  —  64. 


x6  +  6  x5-  40  x3  +  96  x  -  64 


(*2) 


2\3 x6 


x2  +  2x 


3  a;4 1  6  x5 


(x2  +  2  x)3  =  x6  +6  x5  +  12x*+Sscia 
Sx4   -12*4 


(x2  +  2x-4)3  =  x6  +  6  x5  -  40  x3  +  96  x  -  64 

Ans.  x-  +  2  x  —  4. 

2.  Extract  the  cube  root  of  ??i6  —  6  m5  +  40  m3  —  90  m  —  64. 

3.  Extract  the  square  root  of  ai—2a3x  +  3a2x2—2ax3  +  xi. 


176  ALGEBRA. 

4.  Extract  the  fifth  root  of  32  x5  -  80  x*  +  SO  xs  —  40  x" 
+  10  a;  —  1. 

5.  Extract  the  fourth  root  of  xs  —  4  a7  +  10  a;6  —  16a;5  +  19 x4 

-  16  a-3  +  10  x1  -  4  x  +  1. 

248.  When  the  index  of  the  required  root  is  a  multiple  of 
two  or  more  numbers,  we  may  obtain  the  root  by  successive 
extractions  of  the  simpler  roots. 

For,  since  (Art.  17),  (  7  a)mn  =  a, 
taking  the  nth.  root  of  both  members,  we  have  (Art.  235), 

Taking  the  mth  root  of  both  members, 

y/  a==  y  (y  a). 

Or,  the  mnth  root  of  a  quantity  is  equal  to  the  mth  root  of 
the  nth  root  of  that  quantity. 

EXAMPLES. 

1.  Extract    the   fourth   root  of    16  xi  -  96  x3  y  +  216  x2  y2 

-  216  x  ys  +  81  y\ 

2.  Extract  the  sixth  root  of  a12  —  6  a10  +  15  a8  —  20  a6 
+  15  a4  -  6  a-  +  1. 

3.  Extract   the   fourth   root   of    m8  —  8  m1  ±  12  w6  +  40  m5 

-  74  m*  - 120  ms  +  108  m2  +  216  wi  +  81. 


XXII.  — THE  THEORY  OF  EXPONENTS. 

249.     In  Art.  17,  we  defined  an  exponent  as  indicating  how 
many  times  a  quantity  was  taken  as  a  factor;  thus 

am  means  ay.  ay,  a to  m  factors. 

Obviously  this  definition  has  no  meaning  unless  the  expo- 
nent is  a  positive  integer;  and  as  fractional  and  negative  ox- 


EXPONENTS.  177 

portents  have   not  been  previously  defined,  we  may  give    to 
them  any  definition  we  please. 

250.  We  found  (Arts.  82,  93,  and  228)  that  when  m  and 
n  were  positive  integers, 

I.     amXan  —  am  +  n. 

am 
II.     —  =  am~n. 
an 

III.     («'")"  =  amn. 

As  it  is  convenient  to  have  all  exponents  follow  the  same 
laws,  as  regards  multiplication,  division,  and  involution,  we 
shall  define  fractional  and  negative  exponents  in  such  a  way 
as  to  make  Ride  I  hold  for  any  values  of  m  and  n.  We  shall 
now  find  what  meanings  must  be  assigned  to  them  in  con- 
sequence. 

3 

251.  To  find  the  meaning  of  a?. 

As  Rule  I  is  to  hold  universalhT,  it  follows  that 

a        a         a  +  a        s 
a2Xa2  =  a2    2  =  a2=a3. 

a 
Hence,   a2  is  such  a  quantity  as  when  multiplied  by  itself 

3 

produces  a3.    Then,  by  the  definition  of  root  (Art.  17),  a2  must 

a 

be  the  square  root  of  a?  ;  or,  a2  =  \J  az. 

Again,  to  find  the  meaning  of  a3. 

i         i        i.        i+i+i        a 
By  Rule  I,  a3  X  a, 3  X  a3  ■=  a3    3    3  =  a3  =  a. 

Hence,  a*  is  such  a  quantity  as  when  taken  3  times  as  a 
factor  produces  a  ;  or,  a3  =  fya. 

252.  We  will  now  consider  the  general  case. 

p 
To  find  the  meaning  of  aq>  p  and  a  being  positive  integers. 


178  ALGEBRA. 

p_       p       p 
By  Kule  I,  aq  X  a9  X  a5  X to  q  factors 

P  P  P  P  yy 

—  +—-i 1- to?  terms  — X<2 

=  aq    q    «  =aq      =ap. 

p 

Hence,  a5  is  such  a  quantity  as  when  taken   q  times  as  a 

p 

factor  produces  ap.     Then  a'1  must  be  the  qi\\  root  of  ap; 
p 

or,  a*  =  yap. 

3  4  5  1 

For  example,  a*  =  \j  az  \  c2  =\J  c5 ;  x3 '  =  y  x  ;  etc., 
and,  conversely,  y'  a5  =  «4  ;  y7  ic  =  x2  ;  y"  m5  =  m,3"  ;  etc. 

EXAMPLES. 

253.     Express  the  following  with  radical  signs  instead  of 
fractional  exponents : 

a     2  x    i 


a 


3.2  c2.  5.  x*y^.       7.  4a5J«       9.  5yT°£T^. 


2.6^.     4.  3«m^.     6,  rn'iA      8.  2c«t/l     10.  3»^c^l 

Express  the  following  with  fractional  exponents  instead  of 
radical  signs : 

11.  yV.  13.   sjn.  15.   3  y/  m5.  17.   ty  a1  ty  a?. 

12.  yV-  14.   yV.  16.   4  ^a10.  18.   v^vV- 

19.   5sjm»%n<.  20.   2avVy>. 

254.     To  find  the  meaning  of  a-3. 
By  Kule  I,  a~3  X  a3  =  a0  =  1,  by  Art.  94. 

Hence,  a  ~  3  =  — - . 

To  find  the  meaning  of  a~2. 
By  Rule  I,  a~*  X  c$  —  a0  =  1. 

Hence,  a.    2  =  — . 


EXPONENTS.  179 

255.     "We  will  now  consider  the  general  case. 

To  find  the  meaning  of  a~s,  s  being  integral  or  fractional. 

By  Eule  I,  a~s  X  as  =  a°  =  1. 

1 

Hence,  a   s  =  —  . 


1  1  -2  1 

For  example,  ar1  =    ~\a  4  —  ^?5  a    3  —  ~  '■>  e*cv 

a' 


a  a* 


1  X2  2-3 

and,  conversely,  — ^  =  «-2 ;  —  =  a;2  a-3 ;  —  =  2  a  * ;  etc. 

We  observe,  in  this  connection,  the  following  important 
principle : 

^4  quantity  may  be  changed  from  the  denominator  of  a  frac- 
tion to  the  numerator,  or  from  the  numerator  to  the  denomi- 
nator, if  the  sign  of  its  exponent  be  changed. 

EXAMPLES. 

256.  Remove  all  powers  from  the  denominators  to  the 
numerators  in  the  following : 


ar      5a;3      2  a;-1  a;      a;2,    a;-2      x~ 

a2      a3  —  1       a4      a5  —  b 


3. 


4. 


x3 

3 

a-4 

+ 

a;2       a;    5 

7  m 
6c-1 

3  m 

3 

7  c* 

4m2-l      3???3  +  2?i 

i      + 
5  c6              2  c"5 

Remove  all  powers  from  the  numerators  to  the  denominators 
in  the  following : 


*o 


_     2a;       3a"2  a  _    a?      a:3      x~2      2x    1 


180  ALGEBRA. 


a6         3a4      5a2  a 

7.  —^--^ — ^  + 


8. 


a;  +  2         5  b  03         7  —  b  a' 


m— x        m3        %    5  2p 


1  —  x2      3x       5x    1       7  x 


K  ™  — 1  7  ^.  —  3  * 


Express  the  following  with  positive  exponents: 

J_  2.  _3 

9.   2x2y2  —  3x~lyi '—  x~iy    r. 

10.  a-15-2  +  2cr3i-4-3fth~l 

_i     _s  _i 

11.  3a;    3y    7—  4  a*  ?/       +  a  y~  • 

12.  a"1  6~2  c8  +  a~2  b~%  e-4  +  as  j-2  c# 

257.  We  obtained  the  meanings  of  fractional  and  negative 
exponents  on  the  supposition  that  Bule  I,  Art.  250,  was  to 
hold  universally.     Hence,  for  any  values  of  m  and  n, 

amx  an  —  am  +  n 

3  2  3_2.  J^ 

For  example,  a2X «~5  =  a2~5  =  a~3 ;  a4  X  «      —  «*    3  =  a   '  5 
a-iXa*  =  a        2  =  a    2  ;  a3  X  a5  =  a3    °  =  aTo  I  etc. 


EXAMPLES. 

Multiply  together  the  following : 

1.  a8  and  a"1.  4.  c8  and  y'c2.  7.  rc  and  n~?. 

2.  a2  and  a"2.  5.  a:"1  and  (far8.       8.  a^  and  af*. 

3.  a- x  and  a~ 5.  6.  m2  and  ^— .  9.  2  c~  *  and  -  3  a  tycs. 


EXPONENTS.  181 


10.  Multiply  A  ^+2a*-3ftHy26  *-4a  3_6a  £&*. 


a%  b~2  +  2  a^  -  3  b2 
2  J~2  _  4  a-J  _  o  a~%  iih 

2  a$  &-1  +  4  a*  6"^  -  6 


Aa*b~*-S  +  12a  H2. 


6  - 12  a  -J  b  2  +  18  a  %b 


2  a$  b-1  -  20        +  18  a   »  6,  ^ws. 

Note.     It  should  be  carefully  remembered,  in  performing  examples  like 
tbe  above,  that  any  quantity  whose  exponent  is  0  is  equal  to  1  (Art.  94). 

Multiply  together  the  following  : 

11.  a2  b~°-  -  2  +  a~2  b2  and  a2  b~2  +  2  +  a~2  b2. 

3.        i   i        ii        a       ,    i        i 

12.  a4  —  a?  b±  +  a4  b'2  -  b*  and  a4  +  b*. 

13.  a~2  -  2  a"1  b  +  b2-  a  b3  and  ars  +  2  or2  b. 

14.  3  ar1  -a~2b-1+  or*  b~2  and  6  a*  b2  +  2  a2  b  +  2a. 

15.  x~sf  -x~2y-2x-1  and  2  x2 y~l  +  2x8!/-2-±xi y~\ 

16.  x% y~*  +  2  +  jc"  *  y*  and  2  a;- *  y4  _  4 A.-  $  yf  +  2  a-- 2  y4. 

17.  2  sc^  —  3  a;^  —  4  +  aT*  and  3  a;'*'  +  x  —  2  x*. 

18.  4  a4  &-1  +  «4  -  3  «~4  6  and  8  a4  &-1  -  2  a~±  -  6  «~4  6. 

258.     To  prove  that  Rule  II  holds  for  all  values  of 'm  and  n. 
By  Rule  I,        am ~n  X  an  =  am~ n + w  =  am. 

Inverting  the  equation,  and  dividing  by  an,  we  have 


—  =  a" 
an 


182  ALGEBRA. 


ft3  cc~ 2 

For  example,  —  =:a3_1=:a2;     — —  =  a~2~3  =  a~5; 


-I         _|  +  2         f        a3  8  +  4         V- 


a 


-  =  a    4  '  "  =  a4  ;  — -  =  a     5  =  a  5  ;  etc. 


EXAMPLES. 

Divide  the  following: 

_i  _4  l  1 

1.  a3  by  a-1.       4.    a    2  by  a     .         7.  x    by -^— g. 

2.  a  by  a3.  5.   rMiyJc5.  8.  5  »  by  2  or1  $  6. 

3.  a^  by  a*        6.   m2  by  tym,-2.        9.  0  a"1  6^  by  -  3  a  6~~*. 

10.    Divide  2  a^  6"1  -20  +  18  a-"""  6  by  J  b~  *  +  2  J  -  3  6*. 


2  J' b-1- 20  + IS  a    ^b 
2a?b~1+4:cfib~^—6 


2     _l  JL  I 

a:i6    »+-2a*  — 3  6' 


2  6    2  -  4  a   *  —  6  a  $  b 2,  ^?*s. 


-4  a*  6   *  — 14  +  18  a  ^6 
-4a^6~^-   8  +  12  »~^ 6^ 

-6  -12  a"*  6^+  18  a~ %b 

-  6  - 12  a~^  62  +  18  <T%  b 


Note  1.  Particular  attention  must  be  given  to  seeing  that  the  dividend 
and  divisor  are  arranged  in  the  same  order  of  powers,  and  that  each  re- 
mainder is  brought  down  in  the  same  order.  It  must  be  remembered  that 
a  zero  exponent  is  greater  than  any  negative  exponent ;  and  that  negative 
exponents  are  the  smaller,  the  greater  their  absolute  value. 

Note  2.  In  dividing  the  first  term  of  the  dividend  or  remainder  by  the 
first  term  of  the  divisor,  it  will  be  found  convenient  to  write  the  quotient 
at  first  in  a  fractional  form;  reducing  the  result  by  the  principles  of  Art. 
258.     Thus,  in  getting  the  first  term  of  the  quotient  in  Ex.  10,  we  divide 

2  a*  b-1  by  a?  b  '2.    Then,  the  result  =     „    _L    =  2  a3    3  b     +  *  =  2  b  * 

(fit   * 


EXPONENTS.  183 

Divide  the  following : 

jl         i 

11.  a  —  b  by  a5  —  bb. 

12.  a-4  +  a~2  b'2  +  b~i  by  a~2  +  a'1  b'1  +  b~2. 

13.  2x-2y2  +  6  +  Sx2i/-2hj2x  +  2x2y-1  +  4tx3y-2. 

14.  2  x?s  y~x  —  2x~%y  +  32  x~2  y3  by  2  +  6  a-"  §  y  +  8  as"  *  ?/. 

15.  cc~3  ?/-5  —  3  x~5  y~n  +  x-1  y~9  by  x~2  y~s  +  x~ 3  ?/-4  —  a;-  4  y-6. 

16.  8  —  10o;-22/J^+2aj-*2/^>"  by  4x~*y%  +  2x~2y*—2x~^y4. 

259.     To  prove  that  Rule  III  holds  for  all  values  of  m 
and  n. 

We  will  consider  three  cases. 

Case  I.     Let  m  have  any  value,  and  n  be  a  positive  in- 
teger. 

Then,  from  the  definition  of  a  positive  integral  exponent, 

(am)n  =  amXamXam to  n  factors 

—  f,m  +  m  +  m ton  terma  __  ~m  n 

Case  II.     Let  m  have  any  value,  and  n  be  a  positive  frac- 

v 
tion,  which  we  will  denote  bv  -  • 


P_         q, 


Then,  (am)n  =  (am)T=  \  (am)p,  by  the  definition  of  Art.  252, 

=  y7^"^     by  Case  I,  Art.  259, 


mp 

=  aJ,  by  Art,  252, 


=  amxq  =  amn. 


Case  III.     Let  m  have  any  value,  and  n   be  a  negative 
quantity,  integral  or  fractional,  which  we  will  denote  by  —  s. 


184  ALGEBRA. 

Then,  (am)n  =  (am)~s  —  — -rs ,  by  the  definition  of  Art.  255, 

(am)s 

= ,  by  Cases  I  and  II,  Art.  259, 


=  a-ms  =  am{-s)  =  amn. 

Thus,  we  have  proved  Kule  III  to  hold  for  all  values  of  m 
and  n. 

For  example,  (a2)3  =  a6 ;  (a"1)5  =  a"5 ;  (a~?s)  *  =  a"*  ; 

(J)$  =  a;  (a*)~*  =  a~*J  (a2)~*  =  a-*;  etc. 


EXAMPLES. 

260.     Find  the  values  of  the  following : 

5'(C" *)">.      10.    f] 


1.    («•)-■      4.   (O-*-      7.  tf(c   2)2.      10.     -    . 


2.  («-2)3.      5.    (e-*)-2*.    8.    tfm8)   *     11.    (^ 

3.  (a3)*.        6.    (,/*)*  9.    (^f)-5.       12.    {(^VY1- 

261.     To  prove  that  (a  £)n  =  an  bn  for  any  value  of  n. 

In  Art.  228  we  showed  the  truth  of  the  theorem  when  n  was 
a  positive  integer. 

Case  I.     Let  n  be  a  positive  fraction,  which  we  will  denote 

v  E        E   E 

by  —  .    We  have  then  to  show  that  (ab)«  =a«  b'i. 
9. 

[(a  bfy  =  (a  b)p,  by  Art.  259. 


p    p 


EXPONENTS.  185 

[at  &*]*  =  (c£y  (b^)q  =  apbP=  (ab)",  by  Art.  228. 

Hence,  [(aJ)fj*=[a?  J?]3. 


Therefore,  (a  fi)  «=  a«  &«. 

Case  II.     Let  w  be  a  negative  quantity,  which  we  will  de- 
note by  —  s.     We  have  then  to  show  that  (a  b)~s  =  a~3  b~3. 

(a  b)-°  =  -y—^-  =  — ; - ,  by  Art.  228  and  Case  I, 
v      J  (a  b)s       as  bs      J 

=  ars  b~s. 


262.  To  find  the  value  of  a  numerical  quantity  affected 
with  a  fractional  exponent. 

1.  Find  the  value  of  8*. 

From  Art.  252,  we  should  have  S^^y'S5;  and  to  find  the 
value  in  this  way,  we  should  raise  8  to  the  fifth  power,  and 
take  the  cube  root  of  the  result. 

A  better  method,  however,  is  as  follows : 

8^  =  (8*)5,  by  Art.  259, 
=  ($&)*  =  2s  =  32;  Ans. 

Note.  Place  the  numerator  of  the  fractional  exponent  as  the  exponent 
of  the  parenthesis,  and  1  divided  by  the  denominator  as  the  exponent  of  the 
cpiantity  within. 

2.  Find  the  value  of  16"*. 


|=  J__    J_  J_         J_  1_ 

~  16*  ~  (IB*)*  =  ^16)5  ~  (±  2>5  ~±32 : 


186 


ALGEBRA. 


EXAMPLES. 

Find  the  values  of  the  following : 

3.27?.        5.1000-*         7.  (-8)1.         9.   ('f^f. 

36'2  x  16   * 

1  7  s  4^  x  9~2 

4.   36*.        6.  9"*  8.    (-27)*       10.    -^T1 


81"      X 16* 


If  the  numerical  quantity  is  not  a  perfect  power  of  the  de- 
gree indicated  by  the  denominator  of  the  fractional  exponent, 
the  first  method  explained  in  Ex.  1,  Art.  262,  is  the  best. 

For  example,  to  find  the  value  of  72,  we  write  it  \/ 7s,  or 
y/  343 ;  and  find  the  square  root  of  343  to  any  desired  degree 
of  accuracy. 

MISCELLANEOUS  EXAMPLES. 
263.     Extract  the  square  roots  of  the  following : 

4  xy  c*de2 

5.  9  x-*  y2  -12x~sy-2  x~2  +  4  x~x  y~x  +  y~\ 

6.  4x*  +  ±x%  y~*  -15x2y~*  -Sx% y~*  +  lGx§ y~\ 

7.  z8  y~%  +  6  -  4  x~%  y%  +  x~3  y%  -  4  x%  y~*. 


V 
Extract  the  cube  roots  of  the  following : 

8.    ab\  9.   -8a;-4/  10. 

11.    $  if  -  12  y^'x-^  6  y$x-2-i/x-3. 


3m2n  * 


ax5 


EXPONENTS.  187 

Reduce  the  following  to  their  simplest  forms : 


12.   ,^„„^. .  15.   ax~v+2z  a2x+v~3z  az. 


Ji  +  2»i+r 


13.    (xa)-b+(x-a)-b.  16.    (^'"x^X^-1)   *. 

ia      (ax+vY      (  a?   \x~v  ,_     r/  _J_\„_»-i_2_ 

Change  the  following  to  the  form  of  entire  quantities : 
18    15^*2         19  X*V2  20  ^2 


Reduce  the  following  to  their  simplest  forms : 


21.  '         ^         22.   ^=^.  23 


Factor  the  following  expressions  : 
24.   9^-12^  +  4.  25.   a^-3a*-88. 


26.  ar2&  + 5  a"1  &*- 66. 


Factor  by  the  method  of  Art.  117  : 

27.   a-b.  28.   a£-&~£.  29.  ar'"y  — 4«*. 

Factor  by  the  method  of  Art.  119  : 
30.    a  — 6.  31.   a +  6.  32.   x~s+8cm^. 


188  ALGEBRA. 


XXIII.  —  RADICALS. 

264.  A  Radical  is  a  root  of  a  quantity,  indicated  by  a 
radical  sign  ;  as,  yfa,  \Jx  +  1,  y  m2—  2  n  +  3. 

When  the  root  indicated  can  be  exactly  obtained,  it  is  called 
a  rational  quantity  ;  and  when  it  cannot  be  exactly  obtained, 
it  is  called  an  irrational  or  surd  quantity. 

265.  The  Degree  of  a  radical  is  denoted  by  the  index  of 

©  / 

the  radical  sign;  thus,  \] a  is   of  the  second  degree;  \x  +  1 
of  the  third  degree. 

Similar  Radicals  are  those  of  the  same  degree,  with  the 

5/- 6, 

same  quantity  under  the  radical  sign ;  as,  \ax  and  1  y  ax. 

266.  Most  problems  in  radicals  depend  for  their  solution 
on  the  following  important  principle  : 

For  any  values  of  n,  a,  and  b,  by  Art.  236, 


V«X  \b  =  \ab. 


REDUCTION  OF  RADICALS. 

TO  REDUCE  RADICALS  OF  DIFFERENT  DEGREES  TO  EQUIVALENT 
RADICALS  OF  THE  SAME  DEGREE. 

267.     1.    Reduce  y1  2,  ^3,  and  ^  5  to  equivalent  radicals  of 
the  same  degree. 

By  Art.  252,  ^2  =  2*  =  2&  =^2"  =  ^64 


{f  3  =  3^  =  3^  =  v^4  =  y'Sl 
^5  =  5*  =  5&  =  y'  53  =  v'  125 


RADICALS.  189 


RULE. 


Express  the  radicals  10 it h  fractional  exponents  ;  reduce  these 
fractions  to  a  common  denominator  •  express  the  resulting 
fractional  exponents  with  radical  signs;  and,  finally,  reduce 
the  quantities  under  the  radical  signs  to  their  simplest  forms. 

Note.  This  method  affords  a  means  of  comparison  of  the  relative  mag- 
nitudes of  two  or  more  radicals  ;  thus,  in  Example  1,  as  y/  125  is  evidently 
greater  than  y/Sl,  and  y'Sl  than  y/64,  hence  A'o  is  greater  than  y/3,  and 
^3  than  y/2. 

EXAMPLES. 

Reduce  the  following  to  equivalent  radicals  of  the  same 
degree : 

2.  s/3,  \f4,  and  ^5.  5.    \f2~^  \J3~b,  and  ^4^ 

94  **  6, 4. 

3.  y  5,  y  6,  and  y  7.  6.    y  a  +  b  and  y  a  —  b. 

.3/ -,    .*/ „       .  /— 5 ;  -■     3/- 


4.    SJxy,  y  x  z,  and  yyz.         7.    y'cr  —  x2  and  sj az  —  Xs. 

8.  Which  is  the  greater,  ^3  or  y'S? 

9.  Which  is  the  greater,  </2  or  y/3? 
10.    Which  is  the  greater,  ^4  or  $5  ? 

TO  REDUCE  RADICALS  TO  THEIR  SIMPLEST  FORMS. 

268.  A  radical  is  in  its  simplest  form  when  the  quantity 
under  the  radical  sign  is  not  a  perfect  power  of  the  degree 
denoted  by  any  factor  of  the  index  of  the  radical,  and  has  no 
factor  which  is  a  perfect  power  of  the  same  degree  as  the 
radical. 

CASE    I. 

269.  When  the  quantity  under  the  radical  sign  is  a  perfect 
power  of  the  degree  denoted  by  some  factor  of  the  index  of  the 
radical. 


190  ALGEBRA. 

a 

1.    Reduce  y  8  to  its  simplest  form. 

EXAMPLES. 

Reduce  the  following  to  their  simplest  forms : 


2.   #9.  4.   ^27.  6.    \/~a^W. 


3.   ^25  a2.  5.   $125  a3 1>9.  7.   <l 25  "' 


CASE    II. 


270.  When  the  quantity  under  the  radical  sign  has  a 
factor  which  is  a  perfect  power  of  the  same  degree  as  the 
radical. 

1.    Reduce  \J  32  to  its  simplest  form. 


V  32  =  V  16x2  =  (Art.  266)  ^16x^2  =  4^2,  Am. 


2.    Reduce  y/54  a4  x  to  its  simplest  form. 

\/5±aix  =  \/2Tasx2ax  =  %  27~a~3  X  ^2~a~x  =  3  a  \/2~a~x~, 

Ans. 

RULE. 

Resolve  the  quantity  under  the  radical  sign  into  two  factors, 
one  of  which  is  the  greatest  perfect  power  of  the  same  degree 
us  the  radical.  Extract  the  required  root  of  this  factor,  and 
prefix  the  result  to  the  indicated  root  of  the  other. 

Note.  In  case  the  greatest  perfect  power  in  the  numerical  part  of  the 
quantity  cannot  be  readily  determined  by  inspection,  it  may  always  be  ob- 
tained by  resolving  the  numerical  quantity  into  its  prime  factors.  Let  it 
be  required,  for  example,  to  reduce  ^  1944  to  its  simplest  form.  1944  = 
2x2x2x3x3x3x3x3  =  28x35.     Hence, 

^1944  -  V/23lT35  =  V22  x  34  x  y/6  =  18  ^6. 


RADICALS.  191 


EXAMPLES. 

Reduce  the  following  to  their  simplest  forms 


3.  ^50.         6.    ^320.  9.    7^63aH5c6. 

4.  3^24.      7.    2^80.  10.    %250x3fz». 

5.  si  12.         8.    \j98asb2.        11.    <Jl8x3yi-27  x*  y 


12.  \/ax2  —  6ax  +  9a.       14.    \/20  a  x1  +  60  a2  x  +  45  a3. 

13.  SJix^-y2)  (x  +  y).         15.    ^192  a465  +  320  a3  64. 

When  the  quantity  under  the  radical  sign  is  a  fraction,  mul- 
tiply both  terms  by  such  a  quantity  as  will  make  the  denomi- 
nator a  perfect  power  of  the  same  degree  as  the  radical.  Then 
proceed  as  before. 

/2 

16.  Reduce  t  /  -  to  its  simplest  form. 

V/i=\/!=V/(H=Y/ixY/6=^G'-te- 

/9 

17.  Reduce  t  /  -  to  its  simplest  form. 

Reduce  the  following  to  their  simplest  forms : 
18. 


19 
20 


192  ALGEBRA. 

TO  REDUCE  A  RATIONAL  QUANTITY  TO  A  RADICAL  FORM. 
271.     1.    Eeduce  3  a;2  to  a  radical  of  the  third  degree. 


3  x2  =  \7  (3  x2)3  =  \]21  x\  Ans. 

RULE. 

Raise  the  quantity  to  the  power  indicated  by  the  given  root, 
and  write  it  under  the  corresponding  radical  sign. 

EXAMPLES. 

Reduce  the  following  to  radicals  of  the  second  degree : 
a      rr  a      3x  x  —  3 

2.    i  a.  3.   -=-  .  4.    a  +  2  x.  5. . 

5  x  —  2 

6.    Reduce  -=-  to  a  radical  of  the  fourth  degree, 
o 

TO    INTRODUCE   THE  COEFFICIENT  OF  A  RADICAL    UNDER   THE 

RADICAL  SIGN. 


272.     1.    Introduce  the  coefficient  of  2  a  y  3  x2  under  the 


radical  sign. 


2  a  V  3a-2  =  \/ 8  a3  X  V  3 x2  =  (Art.  266)  \f  8  a3  X  3  x2  =  y7  24  a3  x% 

Ans. 

RULE. 

Reduce  the  coefficient  to  the  form  of  a  radical  of  the  given 
degree;  multiply  together  the  quantities  under  the  radical 
signs,  and  write  the  product  under  the  given  radical  sign. 

EXAMPLES. 

Introduce  the  coefficients  of  the  following  under  the  radical 
signs : 

2.  3^5-  4.   4a2\/^U.  6.   5c$Ja. 

3.  2^7.  5.   3yTT^.  7.  (x-l)J(?±^). 


RADICALS.  193 

ADDITION   AND   SUBTRACTION   OF   RADICALS. 

273.     1.    Find  the  sum  of  y/18,  \J21,  J  -,  and  12  y/^ 


18 


By  Art.  270,  .    ^18  =  3^2 

v/27=  3^3 


12 


2.    Subtract  ^48  from  ^162. 

By  Art.  270,  ^162  =  3^6 

^48    =2^6 


^6,  ^ws. 


RULE. 


Reduce  each  radical  to  its  simplest  form.  Combine  the 
similar  radicals,  and  indicate  the  addition  or  subtraction  of 
the  dissimilar. 


EXAMPLES. 

Add  together  the  following  radicals  : 

5 


3.  y'S,  ^18,  and  y/50.  6.   ^20,  t/i  and  J 

4.  ^12,  v/48,  and  v/ 108.         7.   y/|y/|>andy^ 

5.  ^16,^/54,  and  v' 128.         8.   t/i'tf^**^ 


2 

27 


72 

3 


194  ALGEBRA. 

Subtract  the  following : 

9.    v/  45  from  ^  245.  10.    J  ~  horn  J 

Simplify  the  following : 


16 
15 


11.  \/243  a  b2  +  V 75  a*  +^3as-  54  a2  b  +  243  a  b2. 

12.  7^27-^75-y/|  +  v/12-y/l-y/l. 

13.  {^16  +  5^54-^250-^/^  +  ^/1  +  ^/^.        . 


15.    V63 «2« -S4:abx  +  2Sb2x  —  ^7d2x  +  42abx  +  63b2x. 


MULTIPLICATION  OF  RADICALS. 

274.     1.    Multiply  sj  2  by  ^  5. 


^2x^5  =  (Art.  266)  ^2  X  5  =  ^10,  Arts. 

2.  Multiply  y'  2  by  y'  3. 

Reducing  to  equivalent  radicals  of  the  same  degree  (Art. 
267),  we  have 

y/2  X  $3  =  $8  X  \j 9  =  ^72,  Am. 


KULE. 

Reduce  the  radicals,  if  necessary,  to  equivalent  ones  of  the 
same  degree.  Multiply  together  the  quantities  under  the  radi- 
cal signs,  and  write  the  product  under  the  common  radical 
sign. 


RADICALS.  195 


EXAMPLES. 

Multiply  together  the  following  : 


3.  y/ 12  and  y/  3.  6.    V  6  a*  and  \/  5  a?. 

4.  ^2  and  y^.  7.    V^i  V^4,  and  V^^6)  ' 

5.  ^axand^bx.  8.    y/ 2,  y/ 5,  and  t/  ^ • 


9.    Multiply  2y/cc  — 3y/?/  by  4y/a:  +  y/?/. 

2  y/cc  —  3  \J  y 
4y/a;+     y/y 


8ic  —  12  \jxy 

+    2v/^7-3  2/ 


8  «  —  10  v/a;  y  —  3  ?/,  ^4ras. 

Note.     It  should  be  remembered  that  to  multiply  a  radical  of  the  second 
degree  by  itself  is  simply  to  remove  the  radical  sigu  ;  thus, 

y/  x  x  y/  x  =  x. 
Multiply  together  the  following  : 
10.   s/x- 2  and  y]x  +  3.  11.   3y/z  -  5  and  7  sjx-  1. 


12.    s/.T  +  l-V/z-land  y'a;  +  1  +  y/x-1    (Art.  106). 


13.  yV  —  1  —  a  and  V  «'2  —  1  +  «• 

14.  \J  x  —  s]  y+  \J  z  and  \j  x  -\-  \J  y  —  \/  z. 

15.  y/2-y/3  +  y/5  and  ^2  +  ^3  +  ^5. 

16.  3y/5-2y/6  +  y/7  and  6  y/5  +  4  y/6  +  2  y/7. 

17.  4y/3-5y/2-2y/5  and  8  y/3  +  10  y/2  -  4  y/5. 


196  ALGEBRA. 

Simplify  the  following : 

Square  the  following  (Arts.  104  and  105)  : 


20.  2^3-^/2.  22.    \/l-a2+a. 

21.  3^8  +  5^3.  23.    ^^b-\{a~+~b. 

DIVISION   OF  RADICALS. 


275.     Since  (Art,  266),  \faXs/b  =  \fab,  it  follows  that 

\a  b  -i-tya  =  y/6. 


RULE. 


Reduce  the  radicals,  if  necessary,  to  equivalent  ones  of  the 
same  degree.  Divide  the  quantities  under  the  radical  signs, 
and  write  the  quotient  under  the  common  radical  sign. 


EXAMPLES. 

8/ 


1.   Divide  ^  15  by  y/ 5. 

Reducing  to  equivalent  radicals  of  the  same  degree,  we  have 

^15-^5  =  ^225-^125  =  ^/11  =  ^/1,  Ans. 

Divide  the  following : 

2.  v' 108  by  v' 18.  6.    s/2by$/3. 

3.  V^O^by  \l~2Z.  ■  7,  % 2  by  ^3. 

4.  v/54byv/6.  8.   s/ 12  hjsj  2. 

5.  S/lT^by  $3~o~.  9.    \/Ta~by$Ta~. 


RADICALS.  197 


INVOLUTION   OF  RADICALS. 

276.     1.    Raise  fy  2  to  the  fourth  power. 

(^ 2)4  =  (2*)4  =  2*  =  f  24  =  ^  16,  Jn». 
2.    Raise  y/  3  to  the  third  power. 

(ft 3)3  =  (3")3  =  3^  =  3*  =  \J  3,  Ans. 

We  observe  that  in  the  first  example  the  quantity  under  the 
radical  sign  is  raised  to  the  required  power ;  while  in  the  sec- 
ond, the  index  of  the  radical  is  divided  by  the  exponent  of 
the  required  power.     Hence  the  following 

RULE. 

If  possible,  divide  the  index  of  the  radical  by  the  exponent 
of  the  required  poioer.  Otherwise,  raise  the  quantity  tinder 
the  radical  sign  to  the  required  poioer. 

Note.     If  the  radical  has  a  coefficient,  it  may  be  involved  separately. 
The  final  result  should  be  reduced  to  its  simplest  form. 

EXAMPLES. 

3.    Raise  y/5  to  the  third  power. 


4.  Square  \J1. 

5.  Find  the  fourth  power  of  4  y  3  x. 

6.  Find  the  sixth  power  of  y/ a2  x. 


7.  Raise  \l  a  —  b  to  the  fourth  power. 

8.  Raise  3  a\bx  to  the  fourth  power. 

9.  Find  the  value  of  (\/ x  +  l)\ 


10.    Find  the  square  of  4  V^2  —  3. 


198  ALGEBRA. 

EVOLUTION   OF   RADICALS. 


277.     1.    Extract  the  square  root  of  y  6  x2. 
\J( y^)  =  (y7^2)*  =  {(6 x*)*} *  =  (6 a;2)*  =  \/~6x~%  Ans. 


2.    Extract  the  cube  root  of  \/27  x3. 

V  (V27^Q  =  (V27V)*=  {v/(3 z)3P~  =  {(3 a>)*}*  =  (3 a')" 
=  y  3  £,   ^4?zs. 

We  observe  that  in  the  first  example  the  index  of  the  radical 
is  multiplied  by  the  index  of  the  required  root ;  while  in  the 
second,  the  required  root  is  taken  of  the  quantity  under  the 
radical  sign.     Hence  the  following 

RULE. 

If  possible,  extract  the  required  root  of  the  quantity  under 
the  radical  sign.  Otherwise,  multiply  the  index  of  the  radical 
by  the  index  of  the  required  root. 

Note.  If  the  radical  has  a  coefficient,  which  is  not  a  perfect  power  of 
the  same  degree  as  the  required  root,  it  should  he  introduced  under  the 
radical  sign  hefore  applying  the  rule.     Thus, 


y(i\lax  )  =  y(\?l6ax  )  =  ylGax. 
The  final  result  should  be  reduced  to  its  simplest  form. 

EXAMPLES. 

3.  Extract  the  square  root  of  y/2. 

4.  Find  the  cube  root  of  y/  8. 

4  

5.  Find  the  cube  root  of  \a  +  b. 


6.  Find  the  square  root  of  \x2  —  2  x  +  1. 

7.  Extract  the  fifth  root  of  y/32. 


8.  Extract  the  cuhe  root  of  y/27. 

9.  Find  the  value  of  ^(3  y/3). 


RADICALS.  199 

5/ 


10.  Find  the  fourth  root  of  \  x*  yvi. 

11.  Find  the  value  of  \J(±\J2). 

TO  REDUCE  A  FRACTION  HAVING  AN  IRRATIONAL 
DENOMINATOR    TO    AN    EQUIVALENT    ONE 
WHOSE  DENOMINATOR  IS  RATIONAL. 

CASE    I. 

278.      When  the  denominator  is  a  monomial. 

2  b 
1.    Reduce  -; —  to  an  equivalent  fraction  whose  denominator 

sj  a 

is  rational. 

Multiplying  both  terms  by  y/  a, 

2b  _2bsja  _2bsja 


\l  a       y/ a\j  a  a 


,  Ans. 


5 
2.    Reduce  ^-^  to  an  equivalent  fraction  whose  denominator 

is  rational. 

Multiplying  both  terms  by  y/  9, 

5_       5^9    _5v^9_5^9 


RULE. 

Multiply  both  terms  of  the  fraction  by  a  radical  of  the  same 
degree  as  the  denominator,  with  such  a  quantity  under  the 
radical  sign  as  will  make  the  denominator  of  the  resulting 
fraction  rational. 


200  ALGEBEA. 


EXAMPLES. 


Reduce  the  following  to  equivalent  fractions  with  rational 
denominators : 


3      3 
3<   ^2" 

4       1 
$2  a 

5      5 
5'    IN- 
CASE   II. 

6-   r2c 
V '27 a2 

279.      When  the  denominator  is  a  binomial,  containing  only 
radicals  of  the  second  degree. 

1.  Reduce  —  — - r^  to  an  equivalent  fraction  whose  denomi- 
nator is  rational. 

Multiplying  both  terms  by  3  —  y/  2, 

10             10(3-^2)           fK  ^  iA^30-10y/2 
3T72=(3  +  V2)(3-^2)  =  ^Art  106> 7  '  ^ 

2.  Reduce  — —-  to  an  equivalent  fraction  whose  de- 
nominator is  rational. 

Multiplying  both  terms  by  ^  5  +  \J  2, 

v/5  +  v/2_(v/5  +  v/2)(y/5  +  v/2)_7  +  2v/10 
v/5-v/2_(v/5-s/2)(v/5  +  v/2)~"      3        ;  AnS' 

RULE. 

Multiply  both  terms  of  the  fraction  by  the  denominator  with 
the  sign  between  its  terms  changed. 

EXAMPLES. 

Reduce  the  following  to  equivalent  fractions  with  rational 
denominators : 


RADICALS.  201 


4  2y/5  +  y/2  s/a  +  x+^a-x 

3<  3TV2*  "   s/5-3^2'  •    y/^r^-y^T 


a; 


,    4-y/3  y/«-y^  19    yV-l-y/^+1 

2-v/3  s/a+s/x  ^at-l+^at+l 

_   VS-V73  o    2+V«  +  1  1Q     ^  +  V/^2-4 

«•     ~777 To  •  "•      ,  •  A«5-      ,  • 

V/2  +  v/3  1-Va+l  »  — VaJ2-4 


y/ft  +  y/5  1Q     «  — V/V— 1         14      y/x  —  4  y/a  —  2 

sja-\jb'  '    a  +  yjd1—!  '    2\lx  +  3\jx  —  2* 

280.  If  the  denominator  is  a  trinomial,  containing  only 
radicals  of  the  second  degree,  by  multiplying  both  terms  of 
the  fraction  by  the  denominator  with  one  of  its  signs  changed, 
we  shall  obtain  a  fraction  which  can  be  reduced  to  an  equiva- 
lent fraction  with  a  rational  denominator  by  the  method  of 
Case  II.     Thus,  to  reduce  the  fraction 

y/2-v/3-v/7 
^2  +  ^3  +  ^' 

Multiplying  both  terms  byy/2  +  y/3  —  y/7, 

v/2-v/3-v/7_(v/2-v/3-v/7)(v/2  +  v/3-v/T)_6-2v14 
l/2  +  y/3  +  y/7~(\/2  +  s/3  +  \/7)(s/2+sJ3-s/7)     2y/6-2 

_3-y/14 

Multiplying  both  terms  by  y/  6  +  1,  we  Lave 

(3-V/14)(v/6  +  l)_3-v/14  +  3y/6-v/84 
(v/6_l)(v/6  +  l)  -  5 

If  the  denominator  is  a  binomial,  containing  radicals  of  any 
degrees  whatever,  it  is  possible  to  reduce  the  fraction  to  an 
equivalent  form  with  a  rational  denominator ;  but  the  process 
is  more  complicated  than  the  preceding  and  rarely  necessary. 


202  ALGEBEA. 

281.  To  find  the  approximate  value  of  a  fraction  whose 
denominator  is  irrational,  reduce  it  to  an  equivalent  fraction 
whose  denominator  is  rational. 

1.    Find  the  value  of  ^ j^  to  three  decimal  places. 

2-J^  =  (Art.  279)  2-±^-2  =  2-±|^  =  1.707,  An*. 

It  will  be  seen  that  the  value  of  the  fraction  is  obtained  in 
this  way  more  easily  than  by  dividing  1  by  2  —  \J  2,  or  its 
value  .586. 

EXAMPLES. 

Find  the  values  to  three  decimal  places  of  the  following : 

2      2  3         3  4     -  5    V/3-V/2 


IMAGINARY   QUANTITIES. 

282.  An  Imaginary  Quantity  is  an  indicated  even  root  of 

a  negative  quantity  ;  as,  y  —  4,  \/  —  a2. 

In  contradistinction,  all  other  quantities,  rational   or   irra- 
tional, are  called  real  quantities. 

283.  All  imaginary  quantities  may  be  expressed  in  one 

common  form,  which  is,  a  real  quantity  multiplied  by   y —  1. 
For  example, 

y/3^  _  yVX(_i)  _  (Art.  266)  y'  a"  X  ^:Z1  =  a  y^l ; 


also,  y/- 2  =  ^2  X(-l)  =  s/2\[^l. 

Hence,  we  may  regard  ^—1  as  a  universal  factor  of  every 
imaginary  quantity,  and  use  it  in  our  investigations  as  the 
only  symbol  of  such  a  quantity. 


KADICALS.  203 

284.  Imaginary  quantities  may  be  added,  subtracted,  and 
divided  the  same  as  other  radicals ;  but  with  regard  to  multi- 
plication, the  usual  rule  requires  some  modification. 


285.  By  Art.  17,  V—  1  means  such  an  expression  as  when 
multiplied  by  itself  produces  —  1 ; 

or,-        (v/^l7  =  -l; 

also,     (y/I^^v/^lTxV^T^-lV3!; 

and,     (V^1)4  =  (V^1)2X(V^31)2=(-1)X(-1)  =  1. 

By  continuing  the  multiplication,  we  should  find 

(V-l)8  =  l;  etc. 
Or,  in  general,  where  n  is  any  positive  integer, 

(v/=ir+W=T;  (V=i)4B+2=-i;  (V=T)4n+8=-V^; 

(V=1)4"+4  =  L 

MULTIPLICATION  OF  IMAGINARY  QUANTITIES. 

286.  1.    Multiply  ^^a^hy  \f^¥. 

V^^X  V17^^  (Art.  283)  a  <f^l  X  ft  y'^1  =  «  ft  (V11*)2 

=  —  aft,  ^4hs. 


2.    Multiply  V-  2  by  y^3. 

V^2Xy/r3=v/2x\/3x(v/::i)2=-^)  il 


??s. 


3.    Multiply  together  ^—  4,  V—  9,  V~  16>  and  V"  25. 

V^  X  V3^  X  S/^IG  x  V/Ir25  =  2x3x4x5x  (v^)* 
=  120(\/~l)4  =  120,  Arts. 


204  ALGEBRA. 

RULE. 

Reduce  all  the  imaginary  quantities  to  the  form  of  a  real 
quantity  multiplied  by  ^—  1.  Multiply  toy  ether  the  real 
quantities,  and  multiply  the  result  by  the  required  power  of 

EXAMPLES. 

Multiply  the  following : 


4.   4^-3  and  2\J-2.  7.   1  +  V-  1  and  1  -  \J- 1. 


5.   _  3  yL_  a  and  4  V-  b.  8.    \J-  a%  V~  b%  and  y—  c2. 


6.   4  +  V-  7  and  8-2^-7.      9.    a  +  \/—  b  and  a  —  \J—b. 


10.   2  V-  3  —  3  V-  2  and  4  V-  3  +  6  V~  2. 


11.    Divide  V-  «  by  V~  *• 


We  should  obtain  the  same  result  by  using  the  rule  of  Art. 
275 ;  hence,  that  rule  applies  to  the  division  of  all  radicals, 
whether  real  or  imaginary. 

Divide  the  following : 


12.  V-  6  by  V-  2. 

13.  ^^12  by  y'-S. 

Simplify  the  following : 


16.    1  +  S^_1.     (Art.  279). 

1-V-l  2-V-2 


14. 

V- 

-5hj^- 

1. 

15. 

tf- 

-  54  by  y7- 

17.  A  +  t 

-2 

18.  ^£0  +  "-^  .     (Art.  154). 
a  —  \'—b      a  +^-b 

19.    Expand  (2 -V^3)a.        20.   Expand  (2  +  3  Y^2)8. 


KADICALS.  205 

QUADRATIC   SURDS. 

287.  A  Quadratic  Surd  is  the  indicated  square  root  of  an 
imperfect  square  ;  as,  \J  3,  ^  x  +  1. 

288.  A  Binomial  Surd  is  a  binomial  in  which  one  or  both 
of  the  terms  are  irrational. 

289.  The  square  root  of  a  rational  quantity  cannot  be 
equal  to  a  rational  quantity  plus  a  quadratic  surd. 

If  possible,  let  y'  a  =  b  +  y/  c 

Squaring  the  equation,  a  =  b2  +  2  b  \j  c  +  c 

or,  2  b  \J  c  =  a  —  b2  —  c 

.        a  —  b2  —  c 
^C  =  ^b— 

that  is,  a  surd  equal  to  a  rational  quantity,  which  is  impossible. 

290.  If  tin'  sum  of  a  rational  quantity  and  a  quadratic 
surd  be  equal  to  the  sum  of  another  rational  quantity  and 
another  quadratic  surd,  the  two  rational  quantities  will  be 
equal,  also  the  two  quadratic  surds. 

That  is,  if  a  +  \J  b  —  c  +  \J  d 

then  a  =  c  and  \Jb  =  \J  d 

For,  if  a  is  not  equal  to  c,  suppose  a  =  c  -f-  x 
then  c-\-x-\-\Jb  =  c  +  \Jd 

or,  x  +  s/  b  =  y/  d 

which  is  impossible  by  Art.  289.     Hence,  a  must  equal  c,  and 
consequently  y/  b  must  equal  y/  d. 


291.     To  prove  that  ifsja  +  ^b  —  sj  x  +  \J  y,  then  V '  a  —  \j b 
=  s/x  —  sfy. 

Squaring  the  equation        \/  a  +  sjb  =  \J  x  +  \J  y, 
we  have  tc  +  \Jb  —  x  +  2  \l~r~y  +  y 

Whence,  by  Art.  290,  a  =  x  +  y  and  >Jb  =  2  \  xy. 


206  ALGEBRA. 

Subtracting  these  two  equations,  we  have 

a~sJb  —  x  —  2sJxy-\-y 

Extracting  the  square  root,  ^ a  —  ^b  =  ^ x  —  ^  y. 

292.     To  extract  the  square  root  of  a  binomial  surd  whose 
first  term  is  rational. 

For  example,  to  extract  the  square  root  of  a  +  \J  b. 

Assume  \J a  +  \fb  =  \/  x  +  sj  y  (1) 

then  (Art.  291),  ^a—sjb  =  ^x  —  \Jy  (2) 

Multiplying  (1)  by  (2),  \ja"-b  =  x-y  (3) 

Squaring  (1),  a  +  \/b  —  x  +  2^xy  +  y 

Whence  (Art.  290),  a  =  x  +  y.  (4) 

Adding  (3)  and  (4),  a  +  \J~aT^b  =  2x,  or  x=l  +  ^f~b- 

Subtracting  (3)  from  (4),  a  —  SJ  a2— b=2y,  or  ?/= ^- . 

Substituting  these  values  of  x  and  y  in  (1)  and  (2), 

^i^=^(l±^EI)  +  v/(»=^E»).    (5) 

EXAMPLES. 

1.    Find  the  square  root  of  3  +  2  ^  2  or  3  +  y/  8. 
Here  a  =  3  and  b  =  8.     Substituting  in  (5),  we  have 


V3T^=v/(3-±^)  +  V/('^^) 

=v/(^)+v/^H2+i>-<- 


RADICALS.  207 

2.   Find  the  square  root  of  6  —  \J  20. 

Here  a  =  6  and  b  =  20.     Substituting  in  (6),  we  have 


293.     Examples  of  this  kind  may  always  he  solved  hy  the 
following  method : 

3.    Extract  the  square  root  of  14  —  4  y/  6. 


y,14_4v/6  =  V/l^-2v/24  =  V/12-2v/24  +  2 

=  (Art.  116)^12-^2  =  2^3-^2,  Arts. 

4.    Extract  the  square  root  of  43  +  15  \J  8. 


V43  +  15  si  8  =  \M3  +  sj  1800  =  ^43  +  2  sj  450 

=  ^25  +  2^450  +  18  =  v/25  +  V/18=:5  +  3v/2>  ^s- 

EULE. 

Reduce  the  surd  term  so  that  its  coefficient  may  he  2.  Sep- 
arate the  rational  term  into  tiro  parts  whose  product  shall  be 
the  quantity  under  the  radical  sign  (see  first  note  on  page  48), 
writing  one  part  he/ore  the  surd  term  and  the  other  part  after 
it.  Extract  the  square  roots  of  these  parts,  and  connect  them 
by  the  sign  of  the  surd  term. 

The  advantage  of  this  method  is  that  it  does  not  require  the 
memorizing  of  formulae  (5)  and  (6). 

EXAMPLES. 

Extract  the  square  roots  of  the  following : 

5.  12  +  2^35.        8.   35  +  10  <J  10.        11.  20-5^12. 

6.  24-2^63.        9.   12  -sj  108.  12.  14  +  3^20. 

7.  16  +  6^7.         10.   8-v/60.  13.  67  -7  si  12. 


208  ALGEBRA. 

Extract  the  square  roots  of  the  following,  using  formula? 
(5)  and  (6),  Art.  292: 


14.   l-12\/-2.        15.   7  +  30V-2.        16.   35-3V/-16. 


17.   2m-2\Jm--n\  18.   x2  +  a  x  -2  \/ax3. 

Extract  the  fourth  roots  of  the  following  : 
19.   193  +  22  y/ 72.         20.   17-12)/ 2.        21.   97-56^3. 

SOLUTION   OF    EQUATIONS    CONTAINING   RADICALS. 

CASE    I. 

294.      Wlien  there  is  only  one  radical  term  in  the  equation. 


1.    Solve  the  equation    v/ar2  —  5  —  x  =  —  1. 

Transposing,  ^  x2  —  5  =  x  —  1 

Squaring,  x2—  5  =  x2  —  2  cc  +  1 

Whence,  x  =  3,  Ans. 

CASE    II. 
295.      When  there  are  two  radical  terms  in  the  equation. 


2.  Solve  the  equation  \]  x  —  \fx  —  3=1. 
Transposing,  \J  x  —  1  =  ^  x  —  3 
Squaring,  x  —  2\Jx  +  l  =  x  —  3 
Transposing  and  uniting,  —  2  y/  x  =  —  4 

or,  \jx  =  2 

Whence,  x  =  4,  Ans. 

CASE    III. 
296.      When  there  are  three  radical  terms  in  the  equation. 

3.  Solve  the  equation  \J x  +  6  +  \J x  +  13  —  v/4a;  +  37  =  0. 


RADICALS.  209 

Transposing,  \J  x  +  6  +  \/x  +  l§  =  \/4:X  +  3T 

Squaring,  x  +  6  +  2  \/cc2  +  19.r  +  T8  +  x  +  13  =  4  x  +  37 


Transposing  and  uniting,     2  y  as2  +  19  a;  +  78  =  2  a;  +  18 


or,  ^+19  a; +  78  =  a: +  9 

Squaring,  a;2  +  19  x  +  78  =  a;'2  +  18  a;  +  81 

Whence,  x  =  3,  ^f «s. 

RULE. 

297.  Transpose  the  terms  of  the  given  equation  so  that  a 
radical  term  may  stand  alone  in  one  member  ;  then  raise  each 
member  to  a  power  of  the  same  degree  as  the  radical. 

If  there  is  still  a  radical  term  remaining,  repeat  the  op- 
eration. 

The  equation  should  he  simplified  as  much  as  possible  hefore 
performing  the  involution. 

Note.  All  the  examples  in  tliis  chapter  reduce  to  simple  equations  ; 
radical  equations,  however,  may  reduce  to  equations  of  the  second  degree, 
for  the  solution  of  which  see  Chapter  XXIV. 

EXAMPLES. 

Solve  the  following  equations  : 

3/ 


4.   V«-8  =  3.       6.   y'3a;  +  4  +  3  =  6.       8.   8-2^ 


x 


5.    V^-3  =  2.        7.    ^3-1-2  =  1.       9.    o~\2x  =  3. 
10.    y/4ar-19-2a-  =  -l.         14.   6  +  ^x  =  \Jl2  + 


11.    ^^-3^+6-1=1-^.  15.    V/-'-32  +  v/a-r=16. 


12.    yx*  —  6x2  +  2  =  x.  16.    ^x  —  3— Va;  +  12=—  3. 


13.   ^  +  ^  +  5  =  5.  17.    \l2x-l+\J2x  +  9  =  8. 


18.    ^3x  +  10-^3x  +  25  =  -3. 


19.    ^x2-3x  +  5-\'x2-5x-2  =  l. 


210  ALGEBEA. 

20.    \Jx2  +  4  x  +  12  +  \jx2  -  12  x  -  20  =  8. 


21.  v^—  ^-3  =  — . 

y  x 

22.  ^3^+^3^+13  =  ^  ==. 

V  o  x  +  13 

23  V7^  — 3_y/g;  — 4 
V/tc  +  T_"y'ic  +  1' 

24  y/^_+38_y/a;  +  28 

\/#  +  6        \J  x  +  4  ' 


25.    ^-1  +  ^  +  4=^4^  +  5. 


26.    yWl  +  V*-2-V/4jc-3  =  0. 


27.    \j2x-3-  ^8^  +  1  + VlS ^-92  =  0. 


28.  y^  -  3  -  V^  -  14  -  y/4  a  -  155  =  0. 

29.  x-  \^ (9  +  x  \/^~~3)  =  3. 

30.  x  +  l  =  \f(l  +  x\/^r+T6). 

31    v5^y/3_v^+_3 
V/2^"-v/2~V/^  +  2' 

32.    y(a*-3a>x  +  x2\/3^~x)  =  a-x. 


XXIV.  — QUADRATIC  EQUATIONS. 

298.  A  Quadratic  Equation,  or  an  equation  of  the  second 

degree  (Art.  164),  is  one  in  which  the  square  is  the  highest 
power  of  the  unknown  quantity  ;  as, 

ax2  =  b,  and  x*  +  8  x  =  20. 

299.  A  Pure  Quadratic  Equation  is  one  which  contains 
only  the  square  of  the  unknown  quantity  ;  as, 

ax-  =  b;  and  a;2  =  400. 


QUADEATIC  EQUATIONS.  211 

Equations  of  this  kind  are  sometimes  called  incomplete  equa- 
tions of  the  second  degree. 

300.  An  Affected  Quadratic  Equation  is  one  which  con- 
tains both  the  square  and  first  power  of  the  unknown  quan- 
tity ;  as, 

x2  +  8  x  =  20  ;  and  a  x2  +  b  x  —  e  =  b  x2  —  a  x  +  d. 

Equations  of  this  kind,  containing  every  power  of  the  un- 
known quantity  from  the  first  to  the  highest  given,  are  some- 
times called  complete  equations. 


PURE   QUADRATIC  EQUATIONS. 

301.  A  pure  quadratic  equation  can  always  he  reduced  to 
the  form 

x2  =  a, 

in  which  a  may  represent  any  quantity,  positive  or  negative, 
integral  or  fractional.     Thus,  in  the  equation 

20a;2      ,K    9      Jv      41      3-5x2 

Clearing  of  fractions,    80  x2  -  12  (5  x2  +  4)  =  41  -  (9  -  15  x2) 
or,  80  x2  -  60  x2  -  48  =  41  -  9  +  15  x2 

Transposing  and  uniting  terms,  5  x2  =  80 

x2  =  16 

which  is  in  the  form  x2  =  a. 

Equations  of  this  kind  have,  therefore,  sometimes  been  de- 
nominated binomial,  or  those  of  two  terms. 

302.  An  equation  of  the  form 

x2  =  a 

may  he  readily  solved  by  taking  the  square  root  of  each  mem- 
ber.    Thus, 

x  =  ±  \Ja, 


212  ALGEBKA. 

where  the  double  sign  is  used,  because  the  square  root  of  a 
quantity  may  be  either  positive  or  negative  (Art.  237). 

Note.     It  may  seem  at  first  as  though  we  ought  to  write  the  douhle  sign 
before  the  square  root  of  each  member,  as  follows  : 

±x  =  ±  y/a. 

We  do  not  omit  the  double  sign  before  the  square  root  of  the  first  member 
because  it  is  incorrect,  but  because  we  obtain  no  new  results  by  consid- 
ering it.     The  equation  ±  x  =  ±  y/  a  can  be  written  in  four  different  ways, 

thus, 

x  =  ^a 

—  x=tfa 

-x—  -  y/« 

where  the  last  two  forms  are  equivalent  to  the  first  two,  and  become  iden- 
tical with  them  on  changing  all  the  signs.  Hence  it  is  sufficient,  in 
extracting  the  square  root  of  both  members  of  an  equation,  to  place  the 
double  sign  before  one  member  only. 

5x2 
303.     1.    Solve  the  equation  3  x2  +  7  =  —r-  +  35. 

Clearing  of  fractions,  12  x2  +  28  =  5  x2  +  140 

Transposing  and  uniting  terms,         7  x2  =  112 

x2  =  16 
Extracting  the  square  root  of  both  members, 

x  =  ±  4,  Ans. 

RULE. 
'Reduce  the  given  equation  to  the  form  x2  =  a,and  then 
extract  the  square  root  of  both  members. 


EXAMPLES 

Solve  the  following  equations  : 
2.  4a;2- 7  =  29.  4. 


3.   5x2  +  5  =  3x2+5o.  5. 


t  X1  — 

■5  = 

Sx 

2 

-11 

5 

8 
"3 

5 

4  +  x 

4 

— 

X 

QUADRATIC  EQUATIONS.  213 

245 


x 


=  5x.  7.   13-V/3a;2+lG  =  6. 


G 


8.  x+^x*  +  3=   ,-z—z 
y  x-  +  6 


9. 


_y/3 


l_y/l_aja      1  +  y/l- 


a- 


.'• 


_     cc2  5  a;2  _    7         2      335 

10,  Y~     +U^2l~X  +~2l' 

11.  2  (x  -  3)  (x  +  3)  =  (x  +  l)2  -  2  x. 


12.   aa;2  +  5  =  c.  13 


x2  —  b      x2  —  a 


AFFECTED  QUADRATIC  EQUATIONS. 

304.     An  affected  quadratic  equation  may  always  be  reduced 
to  the  form 

x2  -\-jpx  =  q, 

where  p  and  q  represent  any  quantities,  positive  or  negative, 
integral  or  fractional.     Thus,  in  the  equation 

3x  —  3      _         3x  —  6 

5  x 5-  =  2  x  H s — 

x  —  0  J 

Clearing  of  fractions, 

10  x  (x  _  3)  _  (6  x  -  6)  =  4  x  (x  -  3)  +  (3  x  -  6)  (a;  -  3) 

or,       10  a-2  -  30  x  -  6  x  +  6  =  4  x2  —  12  a:  +  3  a2  -  15  a;  +  18 

Transposing  and  uniting  terms,  3  a;2  —  9  x  =  12 

Dividing  by  3,  x2  —  3  x  =  4 

which  is  in  the  form  x2  +  p  x  =■  q. 

Equations  of  this  kind  have,  therefore,  sometimes  been  de- 
nominated trinomial,  or  those  of  three  terms. 


214  ALGEBRA. 

305.     Let  it  be  required  to  solve  the  equation 

x2  -\-px  =  q. 

Equations  of  this  kind  are  solved  by  adding  to  both  members 
such  a  quantity  as  will  make  the  first  member  a  perfect  square, 
and  taking  the  square  root  of  the  resulting  equation.  The  pro- 
cess of  adding  such  a  quantity  to  both  sides  as  will  make 
the  first  member  a  perfect  square,  is  termed  Completing  the 
Square. 

In  any  trinomial  square  (Arts.  104  and  105),  the  middle 
term  is  twice  the  product  of  the  square  roots  of  the  extreme 
terms ;  therefore  the  square  root  of  the  last  term  must  be 
equal  to  half  the  second  term  divided  hy  the  square  root  of 
the  first.     Hence  the  square  root  of  the  quantity  which  must 

be  added  to  x~  +  p  x  to  render  it  a  perfect  square,  is  ~-  —-  x, 
or  — .  Adding  to  both  members  the  square  of  ^,  or  ^-,  we  have 
2  ,  _,,  ,  P ,  p  _±q+p 


X<+px  +  -  =  q  +  ^  =  - 
Extracting  the  square  root  of  both  members, 


X  +  P  =  ±\/±<I+P' 


or,  x  =  —7-± 


p      ^iq+jr 


2  -1-  2 


Thus,  there  are  two  values  of  x, 


p      \  4:  q  +  p'2  ■  p       SJ  4  q  +  p2 


x  =  -t>+       —o i  or 


2  '  2        '  2  2 

We  observe  from  the  preceding  investigation  that  the  quan- 
tity to  be  added  to  complete  the  square  is  found  by  taking  half 
the  coefficient  of  cc,  and  squaring  the  result. 

Hence,  for  solving  affected  quadratic  equations,  we  have  the 
following 


QUADKATIC   EQUATIONS.  215 

RULE. 

Reduce  the  equation  to  the  form  x2  +  p  x  =  q. 

Complete  the  square  by  adding  to  both  members  the  square 
of  half  the  coefficient  of  x.  Extract  the  square  root  of  both 
members,  and  solve  the  simple  equation  thus  found. 

1.  Solve  the  equation      x2  —  3  x  =  4. 

Completing  the   square,   by   adding  to  both   members    the 

.3        9 

square  of  -,  or  -, 

o      0         9       .      9      25 
x2  —  3x  +  -  =  4:+  -  =  — 
4  4      4 

■^  3         5 

Extracting  the  square  root,     x  —  ~  =  ±  ^ 

™  3     5 

Transposing,  a;  =  -  ±  - 

Taking  the  upper  sign,  x  =  -r+  k  =  h  —  4. 

Z         ju         Li 

3     5         2 
Taking  the  lower  sign,  x=7>  —  ^  =  —  7,  =  —  1. 

Ll  Li  Li 

Ans.  x  =  £  or  —  1. 
We  may  verify  these  values  as  follows  : 
Putting  x  =  4  in  the  given  equation,  16  —  12  =  4. 
Putting  x  =  -l,  1  +  3  =  4. 

These  results  being  identical,  the  values  of  x  are  verified. 

2.  Solve  the  equation       3  x2  +  8  x  =  —  4. 

8x4 
Dividing  through  by  3  x2  +  -— -=—  - 

o  o 


216  ALGEBRA. 

Completing  the  square,  by  adding  to  both  members  the  square 

,4        16 

0f3'0ry 

2     8^     16  _       4     16_4 

4         2 

Extracting  the  square  root,  x  +  -  =  ±  - 

o  o 

m  4        2 

1  ransposing,  x  =  —  -  ±  - 


rr  i  •       n  4      2 

lakmg  the  upper  sign,  x  =  —  -  +  -  =  —  - 

Taking  the  lower  sign,  x  =  —  ^  —  ^  =  —  ^  =  —  2. 


2 
.4ns.  cc  =  —  -  or  —  2. 
o 

3.    Solve  the  equation     —  3x2  —  7  x  =  -=-. 

o 

7  x         10 
Dividing  through  by  —  3,         x1  +  -~-  =  — — 

Completing  the  square,  by  adding  to  both  members  the  square 

.  7        49 
of  6'°r36' 


X2 

+ 

7x     49         10     49  __  9 
X+36"     ~~9~+36~36 

Extracting  the 

square 

root, 

7_       3 

X  ~t~    „  it    A 

b          b 

Transposing, 

7     3 
X~     6±6 

Whence, 

2               5       A 

x  =  —  q  or  —  k  )  -4«& 

«5                O 

A  SECOND  METHOD  OF  COMPLETING  THE  SQUARE. 

306.     Although  any  affected  quadratic  equation  may  be 
solved  by  the  method  of  Art.  305,   since  its  rule  is  general, 


QUADRATIC   EQUATIONS.  217 

still  it  is  sometimes  rilore  convenient  to  employ  a  second 
method  of  completing  the  square,  known  as  the  "  Hindoo 
Method." 

An  affected  quadratic,  reduced  to  three  terms,  and  cleared 
of  all  fractions,  may  he  reduced  to  the  form 

a  x2  +  b  x  =  c. 
Multiplying  each  term  by  4  a,  we  have 

By  an  operation  similar  to  that  of  Art.  305,  we  may  show 
that  b2  must  be  added  to  both  members,  in  order  that  the  first 
member  may  be  a  perfect  square.     Thus, 

4  a2  x-  +  4  a  b  x  +  b2  =  b2  +  4  a  c 


Extracting  the  square  root,  2  ax  -\-  b  =  ±  \  b2  -\-  4ta  c 


Transposing,  2  ax  =  —  b  ±  \  b2  +  4  a  c 


tv  -a-       i     q                                            -h±\?b2+4ac 
Dividing  by  2  a,  x  = ^ . 

It  will  be  observed  that  the  quantity  necessary  to  complete 
the  square,  is  the  square  of  the  coefficient  of  x  in  the  given 
equation.     Hence  the  following 

RULE. 

Reduce  the  equation  to  the  form  a  x2  +  b  x  =  r. 

Multiply  both  members  of  the  equation  by  four  times  the 
coefficient  of  x2,  and  add  to  each  the  square  of  the  coefficient 
of  x  in  the  given  equation. 

Extract  the  square  root  of  both  members,  and  solve  the  sim- 
ple equation  thus  produced. 

Note.  The  only  advantage  of  this  method  over  the  preceding  is  in 
avoiding  fractions  in  completing  the  square. 

4.    Solve  the  equation    2  x2  —  7  x  =  —  S. 


218  ALGEBRA. 

Multiplying  both  members  by  four  times  2,  or  8, 

16  x2-5Gx  =  -  24 

Adding  to  each  member  the  square  of  7,  or  49, 

16  x2  —  56  x  +  49  =  -  24  +  49  =  25 
Extracting  the  square  root,   4  x  —  7  =  ±  5 
Transposing,  4cc  =  7±5  =  12or2 

Dividing  by  4,  x  =  3  or  - ,  ^4«s. 

307.     This  method  is  usually  to  be  preferred  in  solving 
literal  equations. 

5.    Solve  the  equation      x2  +  (a  —  1)  x  =  a. 
Multiplying  both  members  by  four  times  1,  or  4, 

4  x2  +  4  (a  —  1)  x  =  4  a 

Adding  to  each  member  the  square  of  a  —  1,  or  (a  —  l)2, 

4  x2  +  4  (a  -  1)  x  +  (a  -  l)2  =  4  a  +  (a  -  l)2 

=  a2  +  2  a  +  1  =  (a  +  l)2 
Extracting  the  square  root, 

2  a;  +  (a  —  1)  =  ±  (a  +  1) 
Transposing,  2  cc  =  —  (a  —  1)  ±  (a  +  1) 

Taking  the  upper  sign,  2  x  =  —  (a  —  1)  +  (a  +  1) 

r=  —  «  +  l  +  «+l=2 


or, 

x  =  l. 

Taking  the  lower  sign, 

2x  =  -(a-l)-(a  +  l) 

=  —  a  +  1  —  a  —  1  =  —  2  a 

or, 

x  —  —  a. 

Ans.  x  =  1  or  —  a. 

308.     In  case  the  coefficient  of  x  in  the  given  equation  is 
an  even  number,  the  rule  may  be  modified  as  follows : 


QUADRATIC  EQUATIONS.  219 

Multiply  both  members  of  the  equation  by  the  coefficient  of 
x2,  and  add  to  each  the  square  of  half  the  coefficient  of  x  in 
the  given  equation. 

6.    Solve  the  equation  7  x2  +  4  x  =  51. 

Multiplying  both  members  by  7,  49  x2  +  28  x  =  357 
Adding  to  each  member  the  square  of  2,  or  4, 

49  x2  +  28  x  +  4  =  361 
Extracting  the  square  root,  7  x  +  2  =  ±  19 

Transposing,  7  a;  —  —  2  ±  19  =  17  or  —  21 

17 

Dividing  by  7,  ic  =  —  or  —  3,  ^l»s. 


SOLUTION   OF   QUADRATIC   EQUATIONS   BY  A 

FORMULA. 

309.     In  Art.  30G,  we  showed  that  if  a  x2  +  b  x  =  c,  then 

-b±  \Jb2  +  ±ac  ... 

X  = 2a *  (1) 

We  may  use  this  as  a  formula  for  the  solution  of  quadratic 
equations  as  follows : 

7.    Solve  the  equation     3  x2  +  5  x  =  42. 
Here  a  =  3,  b  =  5,  c  =  42  ;  substituting  these  values  in  (1), 


-5±\/25  +  504 

X  — 

6 

-5±v/529_-5±23 

8.    Solve  the  equation      110  «2  —  21  x  =  —  1. 
Here  a  =  110,  6  =  —  21,  c  =  —  1  j  substituting  in  (1), 


21  ±1^441 -440       21  ±1       1         1       . 
X  =  ~   -22CT        -  =  -220-  =  I6Orll'^ 


220  ALGEBRA. 

Note.     Particular  attention  must  be  paid  to  the  signs  of  the  coefficients 
in  substituting. 

9.    Solve  the  equation,  —  x2  —  6  x  =  8. 
Here  a =  —  1,  b  =  —  6,  c  =  8;  substituting  in  (1), 


6  ±^36- 32      6±2 
x  = —jr = —  =  —  4  or  —  2,  Ans. 


RULE. 

Reduce  the  equation  to  the  form  a  x2  +  b  x  =  c. 

The  value  of  x  is  then  a  fraction,  ivhose  numerator  is  the 
coefficient  of  x  with  its  sign  changed,  plus  or  minus  the  square 
root  of  the  sum  of  the  square  of  said  coefficient,  and  four  times 
the  product  of  the  second  member  by  the  coefficient  of  x'2 ;  and 
whose  denominator  is  twice  the  coefficient  of  x2. 

310.  The  following  equations  may  he  solved  hy  either  of 
the  preceding  methods,  preference  being  given  to  the  one  best 
adapted  to  the  example  considered.  Special  methods  and 
devices  may  also  be  employed  whenever  any  advantage  can 
thereby  be  gained. 

EXAMPLES. 

Solve  the  following  equations  : 

10.  x°-  +  2x  +  7  =  4:2.  16.   26x  +  lox2  =  -7. 

11.  a;2  -  9  x- 22  =  0.  17.   - 40  +  x  =  6 x\ 

12.  x2-Sx  =  -15.  18.   17x  =  2x2-6. 

13.  z2+18*  =  -65.  19.   ^+*=_1. 

t       7       9  f2 

14.  Gx*+7X-3  =  Q.  20.   x=--^-. 

£       o         o 

3x2      °2 

15.  13  z  -  14  =  3  a;'.  21.    ~-^  =  x. 

5         o 


QUADRATIC   EQUATIONS.  221 

22.  ^_-"-_£  =  a  24.    (as-3)(2as  +  l)=4. 

6  Jo 

23.  ^--^  =  -^.         25.   (*+5)O-5)-(llx+l)=0. 

26.  4x(18x-l)  =  (10x-l)2. 

27.  (3a:-5)2-(>  +  2)2  =  -5. 

28.  (as-l)2-(3as  +  8)2=(2as  +  5)2. 

29    2      as_      5  _21 ^_34 

29<   x  +  2-~2-  d7"   5-as      7         7* 

x         x  — 1_3  _ft    cc  +  1      a?  +  3_8 

30-   ^TI— "F""2'  ^+2      x  +  4_3- 

.t         5  —  as      15  on     3  a?2       1  —  8  x      x 

31.     -3 =  -j-  .  OV. 


5  —  x .        x         4'  '  x  —  7          10          5 

5      3z  +  l      1  .n  2as-l         3x         1 

. =  - .  40. 1 — 

x  a;2         4"  x  3  a;  — 12 


cc 


33.    0  =  -— — -.  41.    \/20  +  x-x2  =  2(x-5). 

3x+4     4x+l  T 


34.  _x__— — -  =  0.  42.  as+V5as  +  10  =  8. 
3a;  +  4      7 as-  4 

._     .         35-3as      ,.  AQ  as*-a*  +  7  n 

35.  6  as  H =  44.  43. =  a;  +  — 

a;  ar  +  3  a;  —  1  o 

14  — r  7  3  22 

36.  4 a -=t-,?  =  14  44.  -^ ^=^- 

a;  +  1  x2  —  4     a;  +  2       o 

45.  ^— +-  =  — - -  + 


x*  —  1     3     3  (as  —  1)      a;  +  1 


222  ALGEBRA. 

._    a  +  3      x  —  3      2  a;  —  3 

46.  -s+-    -s  = r- 

a  +  2      x  —  2       x  —  1 

,„    35  +  2      a  — 2      2a +  16 

47.   TH =-=-      -=-, 

a  —  1      a  +  1        a  +  5 

12  +  5  a      2  +  a  1 


12  —  5a         a         1  —  5a 


49. 

\J  X 

+  1 

—  ya- 

-1 

a 

\Jx 

X  ■+ 

4/9 

+  1 

■  V2 

+ v^- 

-1 

"2* 

50. 

(5x4- 

-      V^ 

3)  = 

=  9. 

K1 

a2  +  36  a 

a 

52.  a  ex2  —  b  ex  +  a  d  x  —  b  d. 

53.  a2  -  2  a  a  +  a2  -  &2  =  0. 
2  a  (a  —  a) a 


54 


3a-2a       4 


1  111 

55.  - =  -  +  -  +  -. 

a  +  b  +  x      a      b     a 

56.  (3  a  -  2)  (a  +  5)  -  (a  -  6)  (5  a  -  16)  =301. 

57.  (2  a  +  3)  (3  a  +  4)  =  (8  +  a)  (2  a  +  9). 

58.  (2  a  -  5)2  -  (2  a  - 1)2 =  8  a  -  5  a2  -  5. 

59.  x-  +  b  x  +  c  x  =  (a  +  c)  (a  —  b). 
Sa-x      6  a2  +  a  b  -  2  b2      b2  x 


60.    abx2  + 


c2 


61.    (3  a2  +  b2)  (a2  -  a  +  1)  =  (3  62  +  a2)  (a2  +  a  +  1). 


PROBLEMS.  223 


XXV.  — PROBLEMS 

LEADING  TO  PURE  OR  AFFECTED  QUADRATIC  EQUATIONS 
CONTAINING  BUT  ONE  UNKNOWN  QUANTITY. 

311.  1.  I  bought  a  lot  of  flour  for  $  175  ;  and  the  number 
of  dollars  per  barrel  was  to  the  number  of  barrels,  as  4  to  7. 
How  many  barrels  were  purchased,  and  what  was  the  price  of 
each  ? 

Let  x  =  the  number  of  dollars  per  barrel, 

7  x 
then  -  =  the  number  of  barrels. 

4 

7  x2 
By  the  conditions,  —r-  =  175 

Whence,  x  =  ±  10. 

Only  the  positive  value  is  applicable,  as  the  negative  value 
does  not  answer  to  the  conditions  of  the  problem. 

That  is,  x  =  10,  the  number  of  dollars  per  barrel, 

7  x 
and  -  =  171,  the  number  of  barrels. 

4 

2.  There  is  a  certain  number,  whose  square  increased  by  30, 
is  equal  to  11  times  the  number  itself.     Required  the  number. 

Let  x  —  the  number. 

By  the  conditions,       x2  +  30  =  11  x 
Solving  this  equation,  x  =  5  or  6. 

That  is,  the  number  is  either  5  or  G,  for  each  of  these  values 
satisfies  the  conditions  of  the  problem. 

3.  I  bought  a  watch,  winch  I  sold  for  $  56,  and  thereby 
gained  as  much  per  cent  as  the  watch  cost  me.  Required 
the  amount  paid  for  it. 

Let  x  =  the  amount  paid,  in  dollars. 

Then  x  =  the  gain  per  cent, 

x  X" 

and  — —  X  x  =  — — -  =  the  whole  gain  in  dollars. 


224  ALGEBRA. 


x2 


By  the  conditions,  -^-r—  =  56  —  x 

J  100 

Solving  this  equation,  x  =  40  or  —  140. 

Only  the  positive  value  of  x  is  here  admissible,  as  the  nega- 
tive result  does  not  answer  to  the  conditions  of  the  problem. 
The  cost,  therefore,  was  $  40. 

Note.  When  two  answers  are  found  to  a  problem,  they  should  be  ex- 
amined to  see  whether  they  answer  to  the  conditions  of  the  problem  or  not. 
Only  those  which  answer  to  the  conditions  should  be  retained. 

PROBLEMS. 

4.  I  have  three  square  house-lots,  of  equal  size.  If  I  were 
to  add  193  square  rods  to  their  contents,  they  would  be  equiv- 
alent to  a  square  lot  whose  sides  would  each  measure  25  rods. 
Required  the  length  of  each  side  of  the  three  lots. 

5.  There  are  two  square  fields,  the  larger  of  which  contains 
25,600  square  rods  more  than  the  other,  and  the  ratio  of  their 
sides  is  as  5  to  3.     Required  the  contents  of  each. 

6.  Find  two  numbers  whose  sum  shall  be  15,  and  the  sum 
of  their  squares  117. 

7.  A  person  cut  and  piled  two  ranges  of  wood,  whose  united 
contents  were  26  cords,  for  356  dimes ;  and  the  labor  on  each 
of  them  cost  as  many  dimes  per  cord  as  there  were  cords  in  its 
range.     Required  the  number  of  cords  in  each  range. 

8.  A  grazier  bought  a  certain  number  of  oxen  for  $  240,  and 
having  lost  3,  he  sold  the  remainder  at  $8  a  head  more  than 
they  cost  him,  and  gained  $59.     How  many  did  he  buy  ? 

9.  The  plate  of  a  rectangular  looking-glass  is  18  inches  by 
12,  and  is  to  be  framed  with  a  frame  all  parts  of  which  are  of 
equal  width,  and  whose  area  is  to  be  equal  to  that  of  the  glass. 
Required  the  width  of  the  frame. 

10.  A  merchant  sold  a  quantity  of  flour  for  §.'59,  and  gained 
as  much  per  cent  as  the  flour  cost  him.  What  was  the  cost  of 
the  flour  ? 


PROBLEMS.  225 

11.  There  are  two  numbers  whose  difference  is  9,  and 
whose  sum  multiplied  by  the  greater  is  266.  What  are  the 
numbers  ? 

12.  A  and  B  gained  by  trade  $  1800.  A's  money  was  in 
the  firm  12  months,  and  he  received  for  his  principal  and  gain 
$2600.  B's  money,  which  was  $3000,  was  in  the  firm  16 
months.     What  money  did  A  put  into  the  firm  ? 

13.  A  merchant  bought  a  quantity  of  flour  for  $  72,  and 
found  that  if  he  had  bought  6  barrels  more  for  the  same 
money,  he  would  have  paid  $>  1  less  for  each  barrel.  How 
many  barrel's  did  he  buy,  and  what  was  the  price  of  each  ? 

14.  A  square  courtyard  has  a  gravel-walk  around  it.  The 
side  of  the  court  wants  2  yards  of  being  6  times  the  breadth 
of  the  gravel-walk,  and  the  number  of  square  yards  in  the  walk 
exceeds  the  number  of  yards  in  the  perimeter  of  the  court  by 
164.     Required  the  area  of  the  court. 

15.  My  gross  income  is  $  1000.  After  deducting  a  percent- 
age for  income  tax,  and  then  a  percentage,  less  by  one  than 
that  of  the  income  tax,  from  the  remainder,  the  income  is 
reduced  to  8  912.  Required  the  rate  per  cent  at  which  the 
income  tax  is  charged. 

16.  The  sum  of  the  squares  of  two  consecutive  numbers  is 
113.     What  are  the  numbers  ? 

17.  Find  three  consecutive  numbers  such  that  twice  the 
product  of  the  first  and  third  is  equal  to  the  square  of  the 
second,  increased  by  62. 

18.  I  have  a  rectangular  field  of  corn  which  consists  of  6250 
hills ;  and  the  number  of  hills  in  the  length  exceeds  the  num- 
ber in  the  breadth  by  75.  How  many  hills  are  there  in  the 
length  and  breadth  ? 

19.  A  certain  company  agreed  to  build  a  vessel  for  $  6300  ; 
but,  two  of  their  number  having  died,  those  that  survived  had 
each  to  advance  $  200  more  than  they  otherwise  would  have 
done.     Of  how  many  persons  did  the  company  at  first  consist  ? 


226  ALGEBRA. 

20.  A  detachment  from  an  army  was  marching  in  regular 
column,  with  C  men  more  in  depth  than  in  front;  hut  Avhen 
the  enemy  came  in  sight,  the  front  was  increased  by  870  men, 
and  the  whole  was  thus  drawn  up  in  4  lines.  Required  the 
number  of  men. 

21.  A  has  two  square  gardens,  and  the  side  of  the  one  ex- 
ceeds that  of  the  other  by  4  rods,  while  the  contents  of  both 
are  208  square  rods.  How  many  square  rods  does  the  larger 
garden  contain  more  than  the  smaller  ? 

22.  A  certain  farm  is  a  rectangle,  whose  length  is  twice  its 
breadth;  but  should  it  be  enlarged  20  rods  in  length  and  24 
rods  in  breadth,  its  contents  would  be  doubled.  Of  how  many 
acres  does  the  farm  consist  ? 

23.  A  square  courtyard  has  a  rectangular  gravel-walk 
around  it.  The  side  of  the  court  wants  one  yard  of  being  six 
times  the  breadth  of  the  gravel-walk,  and  the  number  of  square 
yards  in  the  walk  exceeds  the  number  of  yards  in  the  perim- 
eter of  the  court  by  340.  What  is  the  area  of  the  court  and 
width  of  the  walk  ? 

24.  A  merchant  bought  54  bushels  of  wheat,  and  a  certain 
quantity  of  barley.  For  the  former  he  gave  half  as  many 
dimes  per  bushel  as  there  were  bushels  of  barley,  and  for  the 
latter  4  dimes  per  bushel  less.  He  sold  the  mixture  at  $  1  per 
bushel,  and  lost  $  57.60  by  his  bargain.  Required  the  quan- 
tity of  barley,  and  its  price  per  bushel. 

25.  A  lady  wishes  to  purchase  a  carpet  for  each  of  her  square 
parlors ;  the  side  of  one  of  them  is  1  yard  longer  than  the 
other,  and  it  will  require  85  sqxiare  yards  for  both  rooms. 
What  will  it  cost  the  lady  to  carpet  each  of  the  rooms  with 
carpeting  40  inches  wide,  at  81.75  per  yard  ? 

26.  A  man  has  two  square  lots  of  unequal  dimensions,  con- 
taining together  15,025  square  feet.  If  the  lots  were  contigu- 
ous to  each  other,  it  would  require  530  feet  of  fence  to  embrace 
them  in  a  single  enclosure  of  six  sides.  Required  the  area  of 
each  lot. 


QUADRATIC   EQUATIONS.  907 

27.  A  certain  number  consists  of  two  digits,  the  left-hand 
digit  being  twice  the  right-hand  ;  and  if  the  digits  are  inverted, 
the  product  of  the  number  thus  formed,  increased  by  11,  and 
the  original  number,  is  4956.     Find  the  number. 

28.  A  man  travelled  108  miles.  If  he  had  gone  3  miles 
more  an  hour,  he  would  have  performed  the  journey  in  6  hours 
less  time.     How  many  miles  an  hour  did  he  go  ? 

29.  A  cistern  can  be  filled  by  two  pipes  running  together  in 
2  hours  55  minutes.  The  larger  pipe  by  itself  will  fill  it 
sooner  than  the  smaller  by  2  hours.  What  time  will  each  pipe 
separately  take  to  fill  it  ? 

30.  A  set  out  from  C  towards  D,  and  travelled  3  miles  an 
hour.  After  he  had  gone  28  miles,  B  set  out  from  D  towards 
C,  and  went  every  hour  ^  of  the  entire  distance ;  and  after 
he  had  travelled  as  many  hours  as  he  went  miles  in  an  hour, 
he  met  A.     Required  the  distance  from  C  to  D. 

31.  A  courier  proceeds  from  P  to  Q  in  14  hours ;  a  second 
courier  starts  at  the  same  time  from  a  place  10  miles  behind  P, 
and  arrives  at  Q  at  the  same  time  as  the  first  courier.  The 
second  courier  finds  that  he  takes  half  an  hour  less  than  the 
first  to  accomplish  20  miles.     Find  the  distance  from  P  to  Q. 


XXVI.  — EQUATIONS  IN  THE  QUADRATIC 

FORM. 

312.  An  equation  is  in  the  quadratic  form  when  it  is  ex- 
pressed in  three  terms,  two  of  which  contain  the  unknown 
quantity ;  and  of  these  two,  one  has  an  exponent  twice  as 
great  as  the  other.     As, 

x6  —  6  x3  =  16, 

xs  +  x*  =  72, 
(x-  - 1)2  +  3  {x-  - 1)  =  18,  etc. 


228  ALGEBRA. 

313.     The  rules  already  given  for  the  solution  of  quadratics 

will  apply  to  equations  having  the  same  form.     For,  in  the 

equation 

a  x2n  +  b  xn  =  c, 

let  xn  =  y ;  then  x2n  =  if.     Substituting, 

ay2+by  =  c 
Whence,  hy  Art.  309,  we  have 


or,  x 


• 

—  b±  \Jb2  +  kac 

U  — 

2a 

/*n  — 

—  b±\Jb2  +  4tac 

2  a 

from  which  equation  x  may  he  found  by  extracting  the  wth 
root  of  both  members. 

314.     1.    Solve  the  equation    sc4  —  5  x2  =  —  4. 

The  equation  may  be  solved  as  in  Art.  313,  by  representing 
x-  by  y.     A  better  method,  however,  is  the  following : 

Completing  the  square,         x*  —  5.r2  +  -r-  =  —  4=  +  ~r  —  t 

Extracting  the  square  root,       •        x2  —  -  =  ±  - 

Transposing,  a;2  =  7r±-  =  4orl 

Whence,  x  =  ±  2  or  ±1,  Ans. 

2.    Solve  the  equation  x6  —  6  xs  =  16. 

Completing  the  square,     x6  —  6  xa  +  9  =  16  +  9  =  25 
Extracting  the  square  root,  xa  —  3  =  ±  5 

Transposing,  cc3  =  3±5  =  8or  — 2 

Whence,  x  =  2  or  —  \/2,  Ans. 


QUADRATIC  EQUATIONS.  229 

Here,  although  the  equation  is  of  the  sixth  degree,  we  find 
hut  two  roots.  The  equation  in  reality  has  six  roots,  hut  this 
method  fails  to  give  more  than  two.  It  will  he  shown  here- 
after how  to  obtain  the  other  four. 

3.  Solve  the  equation         x  +  4  \Jx  =  21. 
Writing  the  radical  with  a  fractional  exponent, 

x  +  4x^  =  21 
which  is  in  the  quadratic  form. 
Completing  the  square,  cc  +  4^:r  +  4  =  21  +  4  =  25 

Extracting  the  square  root,       \jx  +  2  =  ±  5 

Transposing,  y/x  =  —  2  ±  5  =  3  or  —  7 

Whence,  squaring,  x  =  9  or  49,  Arts. 

7  1 

4.  Solve  the  equation  3  x2  +  x^  =  3104  x*. 

Dividing  hy  x^,  3x%  +  x%  =  3104 

which  is  in  the  quadratic  form. 

Multiplying  hy  four  times  3,  or  12, 

36  x%  +  12  x%  =  37248 
Completing  the  square,  36  x%  +  12  x%  +  1  =  37249 
Extracting  the  square  root,  6  x®  +  1  =  ±  193 

Transposing,  6  x%  =  —  1  ±  193  =  192  or  - 194 

5  97 

Dividing  hy  6,  x*  =  32  or  —  y 

\  /97\i 

Extracting  the  fifth  root,  xb  =  2  or  —  ( jr-J 

Raising  both  members  to  the  sixth  power, 

x  =64  or  (-4)°>  Ana. 


230  ALGEBRA. 

EXAMPLES. 

Solve  the  following  equations  : 

k    t 

5.  .x4  +  4cc2  =  117.  11.   3  a;* -^-  =  -592. 

6.  a-"4- 9  or"2 +  20  =  0.         12.   a3 -a:2  =56. 

7.  a10 +  31  a-5 -10  =  22.         13.   x-2-s/x  =  0. 


8.  81  x-  +  -4  =  82.  14.   x%  +  x*  =  756. 

x- 

9.  *•  +  !??? -14  =  60.  IS.    ^+2=i=V» 

a;2  4  +  ya:  ya: 


3^* 


2 


10.   a;6 +  20  a;3 -10  =  59.  16.  — —  =  ^-. 

x  —  5         20 


17.    Solve  the  equation  (x  -  5)3  -  3  (x  -  5) 2  =  40. 

a     9  9     169 

Completing  the  square,  (a;  —  5)3— 3  (x — 5)  2  +  -  =  40  +  -  =  — — 

a      3         ^3 
Extracting  the  square  root,  (x  —  5)*  —  ■=  —  ±  -=- 

3      3      13 

Transposing,  (a;  —  5) ^  =  -  ±  —  =  8  or  —  5 

Squaring  both  members,  (x  —  5)3  =  64  or  25 

Extracting  the  cube  root,  x  —  5  =  4  or  \J  25 

Whence,  x  =  9  or  5  +  \J  25, 


QUADRATIC   EQUATIONS.  231 

Solve  the  following  equations  : 

18.  (x2-oa:)2-8(x2-5x)=84:. 

19.  (2  x  - 1)2  -  2  (2  x  -  1)  =  15. 

20.  (3  x-  -2)2-  ll(3a;2-2)  +  10  =  0. 

21.  (;rJ-5)2  +  29  03-5)=:96. 

22.    Solve  the  equation  x*  +  10  x3  +  17  x2  —  40  aj  —  84  =  0. 
We  may  write  the  equation  in  the  form 

x*  +  10  x3  +  2ox2  —  8x2-A0x  =  84 

or,  (;z2+5a;)2-8(cr  +  5a-)=84 

Completing  the  square,  (x2+ox)2—8(x2+5x)  +  16  =  100 
Extracting  the  square  root,  (x2  +  5  x)  —  4  =  ±  10 

Transposing,  (a;2  +  5  x)  =  4  ±  10  =  14  or  —  6 

Taking  the  first  value,  we  have  x2  +  5  x  =  14 

Whence  (Art.  309),  x  =  ~5±^5  +  56  =  =M?  =  2  or  -  7. 

Taking  the  second  value,  we  have  x2  +  5  x  =  —  6 

w,  _5±v/25-24     -5±1 

Vv  hence,  a;  = ^ = - =  —  J  or  —  3. 

Ans.  x  =  2,  —  7,  —  2,  or  —  3. 

Note.  In  solving  equations  of  this  form,  our  object  is  to  form  a  perfect 
trinomial  square  with  the  ,r*  and  x3  terms,  and  a  portion  of  the  x2  term. 
By  Art.  305,  we  may  effect  this  by  separating  the  x2  term  into  two  parts,  one 
of  which  shall  be  the  square  of  the  quotient  obtained  by  dividing  the  x3 
term  by  twice  the  square  root  of  the  x*  term. 


232  ALGEBRA. 

Solve  the  following  equations  : 

23.  x*  -12  x3  +  3±x2  +  12cc  =  35. 

24.  xi  +  2x3-2o  x-  -  26  x  +  120  =  0. 

25.  x4  -  6  cc3  -  29  x2  +  114  x  =  80. 

26.  a4  +  14  x3  +  47  x-  -  14  a;  -  48  =  0. 


27.    Solve  the  equation  2  ar  +  \'2  x2  +  1  =  11. 


We  may  write  the  equation,  (2  x2  +  1)  +  y  2  x'2  +  1  =  12 


49 


Completing  the  square,   (2  x2  +  1)  +  V  2  x2  +  1  +  -  =  -j- 


1  7 

Extracting  the  square  root,  V^^+l  +  T^i^ 


_  1      7 

Transposing,  V  2  ^  +  *  =  ~  2±2=r3°r_4 

Squaring,  2  a;2  +  1  =  9  or  16 

Transposing,  2  x2  =  8  or  15 

15 

Dividing  by  2,  a;2  =  4  or  ~^- 

/15 
Whence,  x  =  ±  2 or ± t/  — , ^4?is. 

Note.  In  solving  equations  of  this  form,  add  such  quantities  to  hoth 
members,  that  the  expression  without  the  radical  in  the  first  member  may 
be  the  same  as  that  within,  or  some  multiple  of  it. 

Solve  the  following  equations  : 


28.   2  x2  +  3  x  -  5^2^+3  x  +  9  =  -3. 


29.   x-  -  6  x  +  5  \?x2  -  6  x  +  20  =  46. 


30.  4  z2+  6  \/4  ar  +  12  x  -  2  =  -  3  (1  +  4  as). 


31.    x2  - 10  a:  -  2  yV2  -  10  ^  +  18  +  15  =  0. 


32.    3.r2+15a:-2v/ar"+5a:  +  l  =  2. 


QUADRATIC  EQUATIONS.  233 

XXVII.— SIMULTANEOUS  EQUATIONS 

INVOLVING  QUADRATICS. 

315.  The  most  general  form  of  an  equation  of  the  second 
degree  containing  two  unknown  quantities,  is 

ax2+bxy+cy2  +  dx  +  ey  +  f=  0, 

where  the  coefficients  a,  b,  c,  etc.  represent  any  quantities, 
positive  or  negative,  integral  or  fractional. 

316.  Two  equations  of  the  second  degree  containing  two 
unknown  quantities  will  generally  produce,  hy  elimination,  an 
equation  of  the  fourth  degree  containing  one  unknown  quantity. 
Thus,  if  the  equations  are 

x2  +  y  =  a 
x  +  y2  =  b 

From  the  first,  hy  transposition,  y  =  a  —  x2 ;  substituting  in 
the  second, 

x  +  (a  —  x2)2  =  b 

or,  x*  —  2  a  x2  +  x  +  a2  —  b  =  0 

an  equation  of  the  fourth  degree.  The  rules  for  quadratics 
are,  therefore,  not  sufficient  to  solve  all  simultaneous  equations 
of  the  second  degree. 

In  several  cases,  however,  their  solution  may  he  effected  hy 
means  of  the  ordinary  rules. 

CASE    I. 

317.  WJien  each  equation  is  of  the  form  a  x2  +  b  y2  =  c. 

1.    Solve  the  equations, 

3x2  +  Aij2  =  7Q 
3  y2  -  11  x2^  4 


234  ALGEBRA. 

Multiplying  the  first  equation  by  3,  and  the  second  by  4, 

9  x2  +  12  if  =  228 
12  f-Ux2  =    16 

Subtracting,  53  x2  =  212 

x2  =  ±,  x  =  ±2. 

Substituting  these  values  in  either  given  equation, 
When  x  =  2,  y  =  ±  4. 

When  a?  =  —  2,  y  =  ±-4. 

^4«s.  #  =  2,  ?/  =  ±  4 ;  or,  x  =  —  2,  y  =  ±  4. 

EXAMPLES. 

Solve  the  following  equations  : 

2.  2  a;2  +  y2  =  9  ;  5  x2  +  6  y2  =  26. 

3.  4  a;2  -  3  f  =  -  11 ;  11  x2  +  5  y2  =  301. 

4.  9  x2  +  24  v/2  =  7  ;  72  a;2  - 180  ?f  —  -  37. 

5.  20  xa  -  16  y2  =  179 ;  5  x2  -  336  ?/2  =  24. 

CASE    II. 

318.      When  one  equation  is  of  the  first  degree. 

1.    Solve  the  equations, 

x2  +  y2  =  13 
x  +  y  —  1 

From  the  second,  by  transposition,  y  =  1  —  x      (1) 

Substituting  in  the  first,  x2  +  1  —  2  x  +  x2  =  13 

or,  cc2  —  x  =6 

AVI  /  A     *     ono\  •        1  ±  S/T+2l         1  ±  5         _ 

\\  hence  (Art.  309),    cc  = L- =  — - —  =  3  or  —  2. 

Substituting  these  values  in  (1), 

When  a  =  3,  y  =  l  —  3  —  —  2. 

WThen  x  =  -2,  y  =  l  +  2  =  3. 

Ans.  x  =  3,  y  =  —  2 ;  or,  x  =  —  2,  y  =  3. 


QUADKATIC  EQUATIONS.  235 

In  solving  examines  under  Case  II,  we  find  an  expression 
for  the  value  of  one  of  the  unknown  quantities  in  terms  of  the 
other  from  the  simple  equation,  which  we  substitute  for  that 
quantity  in  the  other  equation,  thus  producing  a  quadratic 
containing  only  one  unknown  quantity,  by  means  of  which 
the  values  of  the  unknown  quantities  are  readily  obtained. 

Note.  Although  some  examples,  in  which  one  equation  is  of  the  first 
degree  (Ex.  1  for  instance),  may  be  solved  by  the  methods  of  the  next  case, 
yet  the  method  of  Case  II  will  be  found  in  general  the  simplest. 

EXAMPLES. 

Solve  the  following  equations  : 

2.   x-\r  y=  —  1;  xy  =  —  56. 
.     3.   x  +  y  =  3  ;  x2  +  y2  =  29. 

4.  xs  —  y3  =  —  37  ;  x  —  y  =  —  1. 

5.  x  —  y  =  -s--;  xy  =  20. 

6.  10cc  +  y  =  3ccy;3/  —  cc  =  2. 

7.  x  —  y  =  5;  xy  =  —  6. 

8.  x3  +  y3  =  9  ;  x  +  y  =  3. 

9.  3x2-2icy  =  15;  2«  +  3y  =  12. 

10.  x  -  y  =  3  ;  x2  +  y"  =  117. 

11.  x  +  y  —  11;  xy  =  18. 

12.  x  —  y  =  6;  x2  +  y*  =  90. 

13.  x3  +  y3  =  152  ;  x  +  y  =  2. 

14.  x2  +  3  x  y  —  xf  =  23  ;  x  +  2  y  =  7. 

15.  x3-y3  =  9S;  x-y  =  2. 

16.  x  +  y  =  -±;  x2  +  y2  =  58. 

CASE     III. 

319.  When  the  given  equations  are  symmetrical  icith 
respect  to  x  and  y. 


236  ALGEBRA. 

1.    Solve  the  equations, 


x"  +  y2  =  68 
x  y  =  16 


Multiplying  the  second  by  2,  2  x  y  =  32 

Adding  this  to  the  first  equation,    x2  +  2  x  y  +  y'2  =  100  (1) 
Subtracting  it  from  the  first  equation, 

x2-2xy  +  y2=36  (2) 

Extracting  the  square  root  of  (1),  x  +  y  =  ±  10  (3) 

Extracting  the  square  root  of  (2),  x  —  y  —  ±  6  (4) 

Equations  (3)  and  (4)  furnish  four  pairs  of  simple  equations, 
x  +  y  =  10  x  +  y  —  10  x  +  y  =  —  10  x  +  y  =  —  10 
x—y—6  x—y=—Q      x—y=6  x—y=—6 


2x  =  16 

2x  =  4, 

2x  =  -A 

2x  =  -16 

x  =  8. 

x  =  2. 

x  =  —  2. 

x  =  —  8. 

y  =  2. 

y  =  8. 

y  =  -S. 

y  =  -2. 

Ans.  x  =  8,  y  =  2;  x  —  2,y  =  S; 

x  =  —  2,  y  =  —  8 ;  or,  x  =  —  8,  y  =  —  2. 

2.    Solve  the  equations, 

a;3  +  ^  =  133 

x2  —  x  y  +  y2  =  19 

Dividing  the  first  equation  by  the  second, 

x  +  y  =  7  (1) 

Squaring  (1),  x2  +  2  x  y  +  y2  =  49  (2) 

Subtracting  the  second  given  equation  from  (2), 

3  x  y  =  30 ;  or,  4  x  y  =  40  (3) 

Subtracting  (3)  from  (2),  x2  -  2  x  y  +  y2  =  9 

Whence,  x  —  y=±3         (4) 


QUADRATIC   EQUATIONS.  237 

Adding  (1)  and  (4),  2  x  =  10  or  4 

Whence,  x  =  5  or  2. 

Substituting  these  values  in  (1), 
When  x  =  5,  y  =  2 

cc  =  2,  y  =  5. 
Ans.  x  —  5,  y  =  2  ;  or,  a;  =  2,  ?/  =  5. 

The  example  might  have  been  solved  by  substituting  the 
value  of  y  derived  from  (1)  in  either  of  the  given  equations, 
as  in  Case  II. 

The  student  will  notice  the  difference  between  Examples  1 
and  2  as  regards  the  arrangement  of  the  last  portion  of  the 
work. 

3.    Solve  the  equations, 

x  +  y  =  20 


Multiplying  the  first  equation  by  2,     2  x1  +  2  y2  =  416  (1) 

Squaring  the  second  equation,        x2  +  2  x  y  +  y2  =  400  (2) 

Subtracting  (2)  from  (1),  x2  -  2  x  y  +  y2  =  16 

Whence,  x  —  y  =  ±  4         (3) 

Adding  the  second  given  equation  and  (3), 

2  x  =  24  or  16 
Whence,  x  =  12  or  8. 

Substituting  these  values  in  (3), 

When  »  =  12,  y  =  8 

x  =  8,  y  =  12. 
J??5.  aj .=  12,  y  =  8  ;  or,  a;  =  8,  y  =  12. 

This  example  is  solved  more  readily  by  the  method  of  Case 
II;  we  solve  it  by  Case  III  merely  to  show  how  equations 
may  be  solved  symmetrically,  when  one  is  of  the  first  degree. 


238  ALGEBRA. 

EXAMPLES. 

Solve  the  following  equations : 

4.  x'2  +  if  =  25-  xy  =  12. 

5.  x*  +  y2  =  S5;  xy  =  42. 

6.  x3  +  y3  =  -19;  x2-xy  +  y2  =  19. 

7.  x3  -  y3  =  -  65  ;  x2  +  x  y  +  y2  =  13. 

8.  o;  +  2/  =  l;a;y  =  —  6. 

9.  cc2  +  y2  =  65  ;  a;  —  y  =  11. 

10.  a?9+y2  =  61;  x  +  y  =  ll. 

11.  a;8  —  y8  =  117;  a;  —  y  =  3. 

Note.  Exs.  8,  9,  10,  and  11  are  to  be  solved  like  Ex.  3,  and  not  by 
the  method  of  Case  II.  In  solving  Ex.  11,  begin  by  dividing  the  first 
equation  by  the  second. 

CASE    IV. 

320.  When  the  equations  are  of  the  second  degree,  and 
homogeneous. 

Note.  Some  examples,  in  which  both  equations  are  of  the  second  de- 
gree and  homogeneous,  are  solved  more  easily  by  the  methods  of  Cases  I 
and  III,  than  by  that  of  Case  IV.  The  method  of  Case  IV  is  to  be  used 
only  when  the  example  can  be  solved  in  no  other  way. 

1.    Solve  the  equations, 

x2  —  xy  =  35 
xy  +  y2  =  18 

Letting  y  =  vx,  we  have 

35 

x2  —  v  x2  =  35,  or  x2  (1  —  v)  =  35  ;  whence,  x2  =  — -  - —     (1) 

18 

v  x2  +  v2  x2  =  18,  or  x2  (v  +  v2)  =  18  ;  whence,  x2  =  -  '■ — r 

v  +  v 


QUADRATIC  EQUATIONS.  239 

Equating  the  values  of  x2,  -z = -= 

10  1  —  vv+v2 

Clearing  of  fractions,  35  v  +  35  v2  =18  —  18  v 

Transposing  and  uniting,  35  v2  +  53  v  =  18 

Whence  (Art.  309), 


-  53  ±  V  -'809  +  2520     -  53  ±  73     2  9 

V  = 70 =        70        =7°r-5 

2 
If  v  =  -  ,  substituting  in  (1),    x2  =  49,  or  x  =  ±  7 

Substituting  in  the  equation  y  =  v  x, 

o 
When  x  =  7,  y=-xT=2 

x  =  -7,  !/  =  ~x-7  =  -2. 
If  v  =  —  ■=,  substituting  in  (1),       x2  =  -7r- ,  ov  x  =  ±  —j-^ 

£>  2  y  2 

Substituting  in  the  equation  y  =  vx, 

«n  5  9       5  9 

men  x=V2'y=~5x72=~72 

5  9  5         9 


^2'^  5'N      V2      V2 

.4ns.  cc  =  7,  ?/  =  2;  £  =  —  7,  y  =  —  2; 

5_  9  j5_  _9_ 

X~^/2,y~     S/2]  °l,X~     \J2,y~s/2' 

Note.  In  using  the  equation  y~v  x,  to  calculate  the  value  of  y  when 
x  has  been  found,  care  should  be  taken  to  use  that  value  of  v  which  n/i* 
used  in  getting  the  particular  value  of  x. 

EXAMPLES. 

Solve  the  following  equations  : 

2.  x2  +  xy  +  ±y2  =  6;  3x2+8y2  =  U. 

3.  6x2-5xy  +  2y2  =  12;  3  x2  +  2  x  y-  3  y2  =  -3. 


240  ALGEBRA. 

4.  x2  +  x  y  =  12  ;  x  y  —  y2  =  2. 

5.  2  y2  -  4  x  y  +  3  x2  =  17  ;  y2  -x2  =  16. 

6.  x2  +  x  y  —  y2  =  1 ;  x2  —  x  y  +  2  y2  =  8. 

7.  2  x2  —  2  a;  y  —  ?/2  =  3;  x2  +  3a;y  +  2/2  =  11. 

321.  We  append  a  few  miscellaneous  examples,  for  the 
solution  of  which  no  general  rules  can  be  given.  Various  arti- 
fices are  used  ;  familiarity  with  which  can  only  be  obtained  by 
experience. 

1.    Solve  the  equations, 

xz  —  ys  =  19 

x2  y  —  x  y2  =  6 
Multiplying  the  second  by  3,       3  x2  y  —  3  x  y2  =  18  (1) 

Subtracting  (1)  from  the  first  given  equation, 

xz  —  3  x2  y  +  3  x  y2  —  y3  =  1 
Extracting  the  cube  root,  x  —  y  =  1  (2) 

Transposing,  x  =  1  +  y         (3) 

Dividing  the  second  given  equation  by  (2),  x  y  =  6  (4) 

Substituting  from  (3)  in  (4),  y  (1  +  y)  =  6 

or,  y2  +  y  =  6 


m  -l±V/T+24      -1±5 

Vv  hence,  ?/  — k — i5 —  =  ^  or  —  3. 

Substituting  in  (3), 

When  y  =  2,  x  =  3 

y  =  —  3}  x  =  —  2. 

Ans.  x  =  3,  y  =  2 ;  or,  a;  =  —  2,  ?/  =  —  3. 


QUADRATIC  EQUATIONS.  241 

2.  Solve  the  equations, 

V        * 

x  +  y  =  12 

Let  x  —  u  +  v,  and  y  =  u  —  v. 

Then        x  +  y  =  2  u ;  whence,  2u  —  12,  or  u  =  6. 
From  the  first  given  equation,  xz  +  ys  =  18  x  y 

Substituting  x  =  6  +  v,  and  y  =  6  —  v,  we  have 

(6  +  vf  +  (6  -  v)3  =  18  (6  +  v)  (6  - 1>) 
Reducing,  432  +  36  v2  =  648  - 18  v2 

Whence,  54v2  =  216 

v2  =  4,  v  =  ±  2 
Then  a5  =  6+v  =  6±2  =  8or4. 

Substituting  these  values  in  the  second  given  equation, 

When  x  =  8,  y  =  4 

a;  =  4,  ?/  =  8,  ^4ws. 

3.  Solve  the  equations, 

x2  +  y"  +  x  +  y  =  18 

x  ?/  =  6 

Adding  twice  the  second  equation  to  the  first, 

x2  +  2  x  y  +  y2  +  x  +  y  —  30 
or,  (x  +  y)2+(;x  +  y)=30 

N      -  1  ±  Vl  +  120      -1±11        r  r 

Whence,  (x  +  y)  = ^ = „ =  5  or  —  6. 

Taking  the  first  value,  x  +  y  =  5  (1) 

and  the  second  given  equation,  xy  =  6  (2) 

From  (1),  y=5—x  ;  substituting  in  (2),    x2  —  5  x  =  —  6 

5±\/25-24     5±1 
Whence,  x  = —^ =  — „ —  =  3  or  2. 

Substituting  in  (1), 

When,  x  =  3,  y  —  2 

cc  =  2,  y  =  S. 


242  ALGEBRA. 

Taking  the  second  value,  x  +  y  =  —  6  (3) 

and  the  second  given  equation,  x  y  =  6  (4) 

From  (3),  y  —  —  C  —  x  ;  substituting  in  (4), 

x2  +  6  x  =  —  6 

_6±y/36-24      -6+2^3         0       /0 
Whence,  a;  = ^_ = =r_VL_  —  _  3  ±  ^  3. 

Substituting  in  (3), 

When  a=  —  3  +  v/3;  t/  =  —  S-—^3. 

x  =  -3-sJ3,y  =  —  3  +  sJ3. 
Ans.  x  =  3,  y  =  2 ;  cc  =  2,  ?/  =  3  ; 

a;  =  — 3+\/3,  y=— 3— ^3;  or,  x=— 3— y/3,  ?/=— 3  +  y/3 

4.    Solve  the  equations, 

x*  +  7/  =  97       (1) 

*  +y  =~1     (2) 

Raising  (2)  to  the  fourth  power, 

xi  +  4  xs  y  +  6  x2  y2  +  4  x  y3  +  y4  =  1         (3) 
Subtracting  (1)  from  (3),    4  a?3  y  +  6  ar  ?/2  +  4  a;  ?/3  =  —  96 
or,  3  a;2  .?/2  +  2  a;  ?/ (x2  +  ?/2)  =  -  48    (4) 

But  from  (2),  x2  +  y2  =  l-2xy 

Substituting  in  (4),         3  x2  y2  +  2  x  y  (1  -  2  x  y)  =  -  48 
or,  a:2  y2  —  2  cc  y  =  48 

Whence,  x  y  = =  — — —  =  —  6  or  8. 

Taking  the  first  value,  x  y  =  —  6 

From  (2),  y  =  —  1  —  x  ;  substituting,  x2  +  x  =  6 

Whence,  x  = j- —  =  — - =  2  or  —  3. 

Substituting  in  (2), 

When  x  =  2,  y  =  —  3. 

x  =  —  3,  y  =  2. 

Taking  the  second  value,  x  y  =  8 


QUADRATIC  EQUATIONS.  243 

From  (2),    y  =  —  1  —  x ;  substituting,  x2  +  x  =  —  8 

-l±y/T^32    -l±y/-3i 

\\  hence,  x  = -= = „ . 

Substituting  in  (2), 

-  1  +  V/=731           -  1  -  V^3l 
When  x  = ^ >  V~ <j ' 

- 1  -  y/^31  - 1  +  \/^31 

*= — 2 — ->y=-  —2 — 

Ans.  x  =  2,y  =  ~3;  x--S,y  =  2;   x=-1+^~31^ 

_1_S/Z3i                _i_y/Z3i         _i+v/I73i 
y  = y — iov>x  = 2~   ->y= 2 ' 

EXAMPLES. 

Solve  the  following  equations  : 

5.  x  +  y  =  9;  \/x+\/y  =  3.. 

6.  x  +  \J~x~y  +  y  =  19 ;  x2  +  x  y  +  f  =  133. 

7.  a;2  y  +  ar  y2  =  30 ;  x*  if  +  x~yi  =  468. 

8.  x2  +  y2  —  'x  -  y  =  18  ;  x  y  +  x  +  y  =  19. 

9.  x2  +  3x  +  y  =  73  -  2  x  y ;  y2  +  3  y  +  x  =  44. 

. «      n        „      5  x  1/  xy 

10.  as»  +  If=_£j  x-y  =  -f. 

♦  J  4 

„,     a;      4v/#      33  „ 

11.  -  +  -7—  =  -r ;  »  —  y=& 

y     \jy      4 

12-   2  +  8  =  li5  +  S=-4 

13.  x2  y  +  x  y2  =  30 ;  a3  +  ys  =  35. 

14.  cc+V/*'?/=3;?/+V«2/  =  —  2 

15.  x2y  +  y2x  =  6;  -  +  -  =  -. 

16.  r4  +  ?/4  =  17;  x  —  y  =  3. 

17.  z5  -  y5  =  -  211 ;  cc  —  y  =  —  1. 

18.  ar  +  y2  =  7  +  as  y ;  xs  +  ys  =  6  x  y  —  1. 

19.  2x2-7x?/-2?/'2=:5;  3x  y-x2+  6  y2  =  U. 


244  ALGEBRA. 

20.  -A^  +  Jl^S    |  +  |  =  2. 
y  +  3      x  +  J      2    2      o 

21.  z  +  2  =  7;  2y-3*  =  -5;  a;2  +  y1  - z~  =  11. 

22.  xz  =  y2;  (x  +  y)(z—x—y)  =  3;  (x  +  y  +  z)  (z— x—  y)  =  7. 


XXVIII.  —  PROBLEMS 

LEADING   TO    SIMULTANEOUS   EQUATIONS   INVOLVING 

QUADRATICS. 

322.     1.    What  two  quantities  are  those,  the  sum  of  whose 
squares  is  130,  and  the  difference  of  whose  squares  is  32  ? 

Let  x  =  one  number, 

and  y  =  the  other. 

By  the  conditions,  x2  +  y2  =  130 

x2-y2  =  32 
Solving  these  equations,  as  in  Case  I,  Art.  317, 

x  =  9,  y  =  ±  7 ;  # 

or,  x  =  —  9,  y  =  ±  7. 

This  indicates  four  answers  to  the  problem  : 

9  and  7, 
9  and  —  7, 

—  9  and  7, 

—  9  and  —  7. 

Any  one  of  these  pairs  of  values  will  satisfy  the  conditions 
of  the  problem. 

2.   A  says  to  B,  "  The   sum  of  our  money  is  $  18."     B  re- 
plies, "If  twice  the  number  of  your  dollars  were  multiplied  by 


PROBLEMS.  245 

mine,  the  product  would  be  $  154."     How  many  dollars  had 
each  ? 

Let  x  =  A's  dollars, 

and  y  =  B's. 

By  the  conditions,       x  +  y  =  18 

2  x  y  =  154 

Solving  these  equations,  as  in  Case  II,  Art.  318, 

x  =  li  y  =  U; 

or,  x  =  11,  y  =  7. 

That  is,  either  Alias  $7,  and  B  $11,  or  A  has  $11,  and 

B  $7. 

3.  The  price  of  two  coats  and  one  vest  is  $  38.  And  the 
price  of  a  coat  less  that  of  a  vest,  is  to  $  23,  as  $  7  is  to  the 
sum  of  the  prices  of  a  coat  and  vest.  What  is  the  price  of  a 
coat,  and  what  of  a  vest  ? 

Let  x  =  the  price  of  a  coat  in  dollars, 

and  y  =  the  price  of  a  vest. 

By  the  conditions,  2  x  +  y  =  38 

and  x  —  y  :  23  =  7  :  x  +  y 

or  (Art.  181),  a2-?/2  =  161 

Solving  these  equations,  as  in  Case  II,  Art.  318, 

x  =  15,  y  =  8  ; 

107  100 

x  =  —  ,y  =  -  —  . 

Only  the  first  answer  is  admissible,  as  a  negative  value  of 
either  unknown  quantity  does  not  answer  to  the  conditions 
of  the  problem.  Hence,  the  price  of  a  coat  is  $15,  and  of  a 
vest,  $  8. 

Note.  The  note  after  Ex.  3,  Art.  311,  applies  with  equal  force  to  the 
problems  in  this  chapter. 


246  ALGEBRA. 

PROBLEMS. 

4.  The  difference  of  two  quantities  is  5,  and  the  sum  of 
their  squares  is  193.     What  are  the  quantities  ? 

5.  There  are  two  quantities  whose  product  is  77,  and  the 
difference  of  whose  squares  is  to  the  square  of  their  difference 
as  9  to  2.     Required  the  quantities. 

6.  A  and  B  have  each  a  field,  in  the  shape  of  an  exact 
square,  and  it  requires  200  rods  of  fence  to  enclose  hoth.  The 
contents  of  these  fields  are  1300  square  rods.  What  is  the 
value  of  each  at  $  2.25  per  square  rod  ? 

7.  Two  gentlemen,  A  and  B,  were  speaking  of  their  ages. 
A  said  that  the  product  of  their  ages  was  750.  B  replied,  that 
if  his  age  were  increased  7  years,  and  A's  were  diminished  2 
years,  their  product  would  be  851.     Required  their  ages. 

8.  A  certain  garden  is  a  rectangle,  and  contains  15,000 
square  yards,  exclusive  of  a  walk,  7  yards  wide,  which  sur- 
rounds it,  and  contains  3696  square  yards.  Required  the 
length  and  breadth  of  the  garden. 

9.  What  ktwo  numbers  are  those  whose  difference  multi- 
plied by  the  less  produces  42,  and  by  their  sum,  133  ? 

10.  A  and  B  lay  out  money  on  speculation.  The  amount 
of  A's  stock  and  gain  is  $  27,  and  he  gains  as  much  per  cent 
on  his  stock  as  B  lays  out.  B's  gain  is  $  32  ;  and  it  appears 
that  A  gains  twice  as  much  per  cent  as  B.  Required  the  capi- 
tal of  each. 

11.  I  bought  sugar  at  such  a  rate,  that  the  price  of  a  pound 
was  to  the  number  of  pounds  as  4  to  5.  If  the  cost  of  the 
whole  had  been  45  cents  more,  the  number  of  pounds  would 
have  been  to  the  price  of  a  pound  as  4  to  5.  How  many  pounds 
were  bought,  and  what  was  the  price  per  pound  ? 

12.  A  and  B  engage  in  speculation.  A  disposes  of  his  share 
for  $  11,  and  gains  as  many  per  cent  as  B  invested  dollars. 


PROBLEMS.  247 

B's  gain  was  $  36,  and  the  gain  upon  A's  investment  was  4 
times  as  many  per  cent  as  upon  B's.  How  much  did  each 
invest  ? 

13.  A  man  bought  10  ducks  and  12  turkeys  for  $  22.50.  He 
bought  4  more  ducks  for  $  6,  than  turkeys  for  $  5.  What  was 
the  price  of  each  ? 

14.  A  man  purchased  a  farm  in  the  form  of  a  rectangle, 
whose  length  was  4  times  its  breadth.  It  cost  £  as  many  dol- 
lars per  acre  as  the  field  was  rods  in  length,  and  the  number 
of  dollars  paid  for  the  farm  was  4  times  the  number  of  rods 
round  it.  Required  the  price  of  the  farm,  and  its  length  and 
breadth. 

15.  I  have  two  cubic  blocks  of  marble,  whose  united  lengths 
are  20  inches,  and  contents  2240  cubic  inches.  Required  the 
surface  of  each. 

16.  A's  and  B's  shares  in  a  speculation  altogether  amount 
to  $  500.  They  sell  out  at  par,  A  at  the  end  of  2  years,  B  of 
8,  and  each  receives  in  capital  and  profits  $  297.  How  much 
did  each  embark  ? 

17.  A  person  has  $  1300,  which  he  divides  into  two  portions, 
and  loans  at  different  rates  of  interest,  so  that  the  two  por- 
tions produce  equal  returns.  Jf  the  first  portion  had  been 
loaned  at  the  second  rate  of  interest,  it  would  have  produced 
$  36 ;  and  if  the  second  portion  had  been  loaned  at  the  first 
rate  of  interest,  it  would  have  produced  $49.  Required  the 
rates  of  interest. 

18.  Two  men,  A  and  B,  bought  a  farm  of  104  acres,  for 
which  they  paid  $320  each.  On  dividing  the  land,  A  says  to 
B,  "If  you  will  let  me  have  my  portion  in  the  situation  which 
I  shall  choose,  you  shall  have  so  much  more  land  than  I,  that 
mine  shall  cost  $  3  per  acre  more  than  yours."  B  accepted  the 
proposal.  How  much  land  did  each  have,  and  what  was  the 
price  of  each  per  acre  ? 


248  ALGEBRA. 

19.  A  and  B  start  at  the  same  time  from  two  distant  towns. 
At  the  end  of  7  daj's,  A  is  nearer  to  the  half-way  house  than 
B  is,  by  5  miles  more  than  A's  day's  journey.  At  the  end  of 
10  days  they  have  passed  the  half-way  house,  and  are  distant 
from  each  other  100  miles.  Now  it  will  take  B  3  days  longer 
to  perform  the  whole  journey  than  it  will  A.  Required  the 
distance  of  the  towns,  and  the  rate  of  walking  of  A  and  B. 

20.  Divide  the  number  4  into  two  such  parts  that  the  prod- 
uct of  their  squares  shall  be  9. 

21.  The  fore-wheel  of  a  carriage  makes  15  revolutions  more 
than  the  hind-wheel  in  going  180  yards ;  but  if  the  circumfer- 
ence of  each  wheel  were  increased  by  3  feet,  the  fore-wheel 
would  only  make  9  revolutions  more  than  the  hind-wheel  in 
going  the  same  distance.  Find  the  circumference  of  each 
wheel. 

22.  A  ladder,  whose  foot  rests  in  a  given  position,  just 
reaches  a  window  on  one  side  of  a  street,  and  when  turned 
about  its  foot,  just  reaches  a  window  on  the  other  side.  If 
the  two  positions  of  the  ladder  are  at  right  angles  to  each 
other,  and  the  heights  of  the  windows  are  36  and  27  feet  re- 
spectively, find  the  width  of  the  street  and  the  length  of  the 
ladder. 

23.  A  and  B  engaged  to  reap  a  field  for  90  shillings.  A 
could  reap  it  in  9  days,  and  they  promised  to  complete  it  in  5 
days.  They  found,  however,  that  they  were  obliged  to  call  in 
C,  an  inferior  workman,  to  assist  them  the  last  two  days,  in 
consequence  of  which  B  received  3.?.  9d.  less  than  he  other- 
wise would  have  done.  In  what  time  could  B  and  C  each  reap 
the  field? 

24.  Cloth,  being  wetted,  shrinks  J  in  its  length  and  ,\;  in  its- 
width.  If  the  surface  of  a  piece  of  cloth  is  diminished  by  5^ 
square  yards,  and  the  length  of  the  four  sides  by  4}  yards, 
what  was  the  length  and  width  of  the  doth  originally? 


THEORY  OF  QUADRATIC  EQUATIONS.  249 


XXIX.  — THEORY  OF  QUADRATIC  EQUA- 
TIONS. 

323.  A  quadratic  equation  cannot  have  more  than  two 
roots. 

We  have  seen  (Art.  304)  that  every  complete  quadratic 
equation  can  be  reduced  to  the  form 

x2  +  p  x  =  q. 

Suppose,  if  possible,  that  a  quadratic  equation  can  have  three 
roots,  and  that  r1;  r2,  and  r3  are  the  roots  of  the  equation 
x°-  +  p  x  —  q.     Then  (Art.  166), 

r12+pr1  =  q  (1) 

r22+pr2  —  q  (2) 

r32+pr3  —  q  (3) 

Subtracting  (2)  from  (1),     (?y  —  r.F)  +  p  (rx  —  r2)  =  0 

Dividing  through  by  ry  —  r2,  which  by  supposition  is  not 
zero,  as  the  roots  are  not  equal, 

r\  +  r2+p  =  0 

Similarly,  by  subtracting  (3)  from  (1),  we  have 

n  +  n + p  —  o 

Hence,  rx  +  r2  +  p  =  rx  +  rs  -f  p 

or,  r2  =  r3. 

That  is,  two  of  the  roots  are  identical.  Therefore,  a  quad- 
ratic equation  cannot  have  more  than  two  roots. 

DISCUSSION   OF  THE   GENERAL  EQUATION. 

324.  By  Art.  305,  the  roots  of  the  equation  x2  +  p  x  =  q 


are 


-p-\-  \/p2+  ±q         1—p  —  \Jp2+4:q 


250  ALGEBRA. 

1.    Suppose  q  positive. 

Since  p2  is  essentially  positive  (Art.  227),  the  quantity 
under  the  radical  sign  is  positive  and  greater  than  ,p° ;  so 
that  the  value  of  the  radical  is  greater  than  P-  Hence,  one 
root  is  positive,  and  the  other  negatnre. 

If  p  is  positive,  the  negative  root  is  numerically  the  larger; 
if  p  is  zero,  the  roots  are  numerically  equal ;  and  if  p  is  nega- 
tive, the  positive  root  is  numerically  the  larger. 

2.    Suppose  q  equal  to  zero. 

The  quantity  under  the  radical  sign  is  now  equal  to  p2 ;  so 
that  the  value  of  the  radical  is  p.  Hence,  one  of  the  roots  is 
equal  to  0.  The  other  root  is  positive  when  p  is  negative,  and 
negative  when  p  is  positive. 

3.  Suppose  q  negative,  and  4  q  <  p2. 

The  quantity  under  the  radical  sign  is  now  positive  and  less 
than  p>2 ;  so  that  the  value  of  the  radical  is  less  than  p. 

If  p  is  positive,  hoth  roots  are  negative ;  and  if  p>  is  nega- 
tive, both  roots  are  positive. 

4.  Suppose  q  negative,  and  4  q  =p2. 

The  quantity  under  the  radical  sign  is  now  equal  to  zero ; 
so  that  the  two  roots  are  equal ;  being  positive  if  p  is  negative, 
and  negative  if  p  is  positive. 

5.  Suppose  q  negative,  and  4  q  >  p2. 

The  quantity  under  the  radical  sign  is  now  negative  ;  hence, 
by  Art.  282,  both  roots  are  imaginary. 

325.     All  these  cases  may  be  readily  verified  by  examples. 

Thus,  in  the  equation  x2  —  3;r  =  70,  as  p  is  negative  and  q 
positive,  we  should  expect  to  find  one  root  positive  and  the 
other  negative,  and  the  positive  root  numerically  the  larger 

And  this  is  actually  the  rase,  for  on  solving  the  equation,  wfc 
find  x  =  10  or  -  7. 


THEORY  OF  QUADRATIC  EQUATIONS.  251 

326.     From  the  quadratic  equation  x2+px  =  q,  denoting 
the  roots  by  rx  and  r.2,  we  have 


-p+S/Y  +  lq    _1m         -p-Sjp'  +  lq 

2 


n  =  — ^ ,  and  r2 


Adding  these  together,  we  have 

2p 
rx  +  r2  =  —  -^-  =  —p. 

Multiplying  them  together,  we  have 

n  r2  =£=M±±1±  (Art.  106)  =  -  *±  =  -  q. 

That  is,  if  a  quadratic  equation  be  reduced  to  the  form 
x2  +  p  x  =  q,  the  algebraic  sum  of  the  roots  is  equal  to  the  co- 
efficient of  the  second  term,  with  its  sign  changed  •  and  the 
product  of  the  roots  is  equal  to  the  second  member,  with  its 
sign  changed. 

327.  The  equation  a  x2  +  b  x  +  c  =  0,  by  transposing  c,  and 
dividing  each  term  by  a,  becomes 


x2-\ 


bx  c 

a  a 


Denoting  the  roots  of  the  equation  by  xl  and  x2,  we  have,  by 
the  previous  article, 

b  _c 

Xi  -j-  Xo  — ,  and  x±  x.y  —  —  • 

a  a 


328.  A  Quadratic  Expression  is  a  trinomial  expression  of 
the  form  a  x2  +  b  x  +  c.  The  principles  of  the  preceding 
article  enable  us  to  resolve  any  quadratic  expression  into  two 
binomial  factors. 

The  expression  a  x'2  +  b  x  +  c  may  be  written 

f  0      bx      c 

a  I  x  -\ 1 — 

V  a       a 


252  ALGEBRA. 

b  c 

By  the  previous  article,  -  =  —(%  +  x2),  and  -  =x1x2,  where 

xx  and  x2  are  the  roots  of  the  equation  ax2jrbx  +  c  =  Q; 
which,  we  ohserve,  may  he  obtained  by  placing  the  given 
expression  equal  to  0.     Hence, 

ax2  +  bx  +  c  =  a  [x2—  (xx  +  x2)  x  +  xx x2~\. 

The  expression  in  the  bracket  may  be  written 

/V»**   rtn    /yt  ry*    /y>  I /y»       sy 

which,  by  Case  II,  Chap.  VIII,  is  equal  to  (x  —  x^  (x  —  x2). 
Therefore,  a  x2  +  b  x  +  c  =  a  (x  —  xx)  (x  —  x2). 

1.    Factor  6  x2  +  11  x  -+-  3. 

Placing  the  expression  equal  to  0,  and  solving  the  equation 
thus  formed,  we  find 


_  - 11  +  ^121  -  72  _  -  11  ±  7  _       3  1 

X~~         ~V1~         ~_~~12       -~2'0r~3- 

Then,  a  =  6,  x1  =  —  ^,  x2  =  —  5. 

Therefore,  6  x2  +  11  x  +  3  =  6  (x  +  |)  (»  +  5) 

=  (2  a;  +  3)  (3  x  +  1),  Ans. 

2.    Factor  4  +  13  x  -  12  x2. 

Placing  the  expression  equal  to  0,  and  solving  the  equation 
formed,  we  have 


_  - 13  ±  y/ 169  +  192  _  -  13  ±  19  _  4  1 

X~  -24  -24      ~3'  °r      4 

4  1 

Then,  a  =  —  12,x1  =  7r,  x2  =  —  -. 


THEORY  OF  QUADRATIC   EQUATIONS.  253 

Therefore,  4  +  13  x  - 12  x2  =  - 12  he  -  ^J  (a;  +  ^J 

=  -3{x-l)4(x  +  \) 

—  (4  —  3  x)  (4  a;  +  1),  ^4?is. 

Note.  It  should  be  remembered,  in  using  the  formula  a  (x-x{)  (x  -  a'2N, 
that  a  represents  the  coefficient  of  x2  in  the  given  expression  J  hence,  in 
Example  2,  we  made  a=  -  12. 

EXAMPLES. 

Factor  the  following  expressions  : 

3.  x1  +  73  x  +  780.  9.  8z2  +  18z-5. 

4.  x2- 11  x  +  18.  10.  4  z2  -  15  *  +  9. 

5.  x2-4cc-60.  11§  2x2+x-6. 

6.  a8 +  10  a; -39.  12.  9x2-12a-  +  l. 

7.  2  a;2 -7  a; -15.  13.  l-8x-x2. 

8.  21  a-2  +  58  a;  +  21.  14.  49  x2  +  14  a-  -  19. 

329.     The  principles  of  Art.  328  furnish  a  method  of  form- 
ing a  quadratic  equation  which  shall  have  any  required  roots. 

For,  the  equation  a  x2  +  b  x  +  c  =  0,  if  its  roots  be  denoted 
by  xx  and  x2,  may  be  written,  by  Art.  328, 


a 


(x  —  Xy)  (x  —  x2)  =  0,  or  (x  —  £Cj)  (x  —  a-2)  =  9. 


Hence,  to  form  an  equation  whose  roots  shall  be  xv  and  x.,, 
we  subtract  each  of  the  two  roots  from  x,  and  place  the  product 
of  the  resulting  binomials  equal  to  zero. 

7 
1.    Required  the  equation  whose  roots  are  4  and  —  -r  • 

By  the  rule,  (x  —  4)  (x  +  -)=  0 


254 


or, 


ALGEBRA. 

x2- 

9x 

-7  = 

=  0 

4x2- 

9x- 

-28  = 

=  o, 

Ans. 

Clearing  of  fractions, 

EXAMPLES. 

Form  the  equations  whose  roots  are 

17 

2.  1  and  —  2.      5.    7  and  —  6  \ .  8.   —  -^-  and  0. 

8         4 

3.  4  and  5.         6.    —  -  and  - .  9.   1  +  sj  5  and  1  —  ^5. 

o  i 

4.  3  and  —■=.     7.   —  2 J  and  —  3^.     10.   m  +  \/  w  and  m  —  \Jn. 

o 

330.  By  Art.  328,  the  equation  a  x2  +  b  x  +  c  =  0  may  be 
written  (x  —  a^)  (x  — x2)  =  0,  if  acj  and  x2  are  its  roots;-  we 
observe  that  the  roots  may  be  obtained  by  placing  the  factors 
of  the  first  member  separately  equal  to  zero,  and  solving  the 
simple  equations  thus  formed. 

This  principle  is  often  useful  in  solving  equations. 

1.  Solve  the  equation      (2x  —  3)  (3  x  +  5)  =  0. 

3 

Placing  the  first  factor  equal  to  zero,  2x  —  3  =  0,  or  a;  =  - . 

A 

5 

Placing  the  second  factor  equal  to  zero,  ox+  5=0,  orx= —  - . 

A  3  5 

Ans.  x  =  -  or  —  ^ . 

A  o 

2.  Solve  the  equation  x'2  +  5  x  =  0. 
The  equation  may  be  written          x  (x  +  5)  =  0 

Placing  the  first  factor  equal  to  zero,  x  =  0. 

Placing  the  second  factor  equal  to  zero,  x  +  5  =  0,  or  x  =  —  5. 

Ans.  x  =  0  or  —  5. 


THEORY   OF  QUADRATIC  EQUATIONS.  255 

EXAMPLES. 

Solve  the  following  equations  : 

3.  (as-|)  (as-2)  =  0.  9.   2as8-18as  =  0. 

4.  (as+.5)(as-l)  =  0.  10.   (2as  +  5)(3as-l)  =  0. 

5.  (as-?)  (as  +  ?)=0.  11.   (aas  +  b)  (cx-d)  =0. 

6.  (as  +  8)(as  +  i)=0.  12.    (x2-4)  (as2-9)  =  0. 

7.  2as2-13as  =  0.  13.    (3 a;  +  1)  (4 x* - 25)  =  0. 

8.  3  as3  +  12  a:2  =  0.  14.    (as2-a)(as2-aas-£)=0. 

15.  as  (2 as  +  5)  (3 x -7)  (4 x  +  1)  =  0. 

16.  (x2-5x  +  6)(x2  +  7x+  12)  (2 x*  +  9x-5)  =  0. 

331.  Many  expressions  may  be  factored  by  the  artifice  of 
completing  the  square,  used  in  connection  with  the  method  of 
Case  IV,  Chapter  VIII. 

1.  Factor  xi  +  a\ 

x*  +  a4  =  x*  +  2  x2  a2  +a*-2  x2  a2 
=  (x2  +  a2)2  -  (a  x  ^/ 2)2 
=  (Art.  117)  (x2  +  a  x  ^  2  +  a2)  (x2  —  a  x  \J  2  +  a2),  Ans. 

2.  Factor  x2  —  ax  +  a2. 

as2  —  ax  +  a2  =  x2  +  2ax  +  a2  —  3ax. 

=  (x  +  a)2-(^3axY 


=  (x  +  V^3  a  x  +  a)  (x  —  \J3  a  x  +  a),  Ans. 


256  ALGEBRA. 

EXAMPLES. 

Factor  the  following  expressions  : 

3.  x2  +  l.  5.   a2-3ab  +  b*.  7.   x2-x-l. 

4.  x2+x+l.         6.   xi-lx2y2-^y\         8.   mi  +  m2n2+?i\ 

332.  We  have  seen  (Art.  330)  that  any  equation  whose 
first  member  can  be  factored,  and  whose  second  member  is 
zero,  may  be  solved  by  placing  the  factors  separately  equal  to 
zero  and  solving  the  equations  thus  formed.  This  method  of 
solution  is  frequently  the  only  one  which  will  give  all  the  roots 
of  the  equation. 

1.    Solve  the  equation     x3  =  1. 

The  equation  may  be  written  x3  —  1  =  0,  or  (Art.  119), 
(x  -  1)  (x2  +  x  +  1)  =  0. 
Placing  the  first  factor  equal  to  zero, 

x  —  1  =  0,  or  x  =  1. 
Placing  the  second  factor  equal  to  zero, 

x2  +  x  +  1  =  0,  or  x2  +  x  =  —  1 


Whence  (Art,  309),  -  -  =  ^  ±  ^ 


x  = 


2  2 


Hence,  x  =  1  or ^1 ,  Ans. 


EXAMPLES. 

Solve  the  following  equations  : 

2.  xi  =  -l.         4.   x4+«4  =  0.  6.   x6  =  l. 

3.  x3  =  -l.         5.   X4-X*+1  =  0.         7.   rr4-  — +1  =  0. 

These  examples  afford  an  illustration  of  the  statement  made 
in  Art.  167  that  the  degree  of  an  equation  indicates  the  num- 
ber of  its  roots. 


DISCUSSION   OF   PROBLEMS.  257 

XXX.  —  DISCUSSION  OF  PROBLEMS 

LEADING  TO  QUADRATIC  EQUATIONS. 

333.  In  the  discussion  of  problems  leading  to  quadratic 
equations,  we  find  involved  the  same  general  principles  which 
have  been  established  in  connection  with  simple  equations 
(Arts.  205-212),  but  with  certain  peculiarities. 

These  peculiarities  will  be  now  considered.  They  arise  from 
two  facts : 

1.  That  every  quadratic  equation  has  two  roots  •  and 

2.  That  these  roots  are  sometimes  imaginary. 

334.  In  the  solution  of  problems  involving  quadratics,  it 
has  been  observed  that  the  positive  root  of  the  equation  is 
usually  the  true  answer ;  and  that,  when  both  roots  are  posi- 
tive, there  may  be  two  answers,  either  of  which  conforms  to 
the  given  conditions. 

The  reason  why  results  are  sometimes  obtained  which  do 
not  apply  to  the  problem  under  consideration,  and  are  there- 
fore not  admissible,  is  that  the  algebraic  mode  of  expression 
is  more  general  than  ordinary  language ;  and  thus  the  equa- 
tion which  conforms  properly  to  the  conditions  of  the  problem 
will  also  apply  to  other  conditions. 

1.  Find  a  number  such  that  twice  its  square  added  to  three 
times  the  number  may  be  65. 

Let  x  =  the  number. 

Then        2  x2  +  3  x  =  65  (1) 

13 

Whence,  x  =  5  or —  . 

The  positive  value  alone  gives  a  solution  to  the  problem  in 
the  sense  in  which  it  is  proposed. 

To  interpret  the  negative  value,  we  observe  that  if  we 
change  x  to  —  x,  in  equation  (1),  the  term  3  x,  only,  changes 


258  ALGEBRA. 

its  sign,  giving  as  a  result  the  equation  2  x'2  —  3  x  =  65.     Solv- 

13 

ing  this  equation,  we  shall  find  x  =  —  or  —  5,  which  values 

only  differ  from  the  others  in  their  signs.     We  therefore  may 

13 

consider  the  negative  solution, — ,  taken  independently  of 

Li 

its  sign,  the  proper  answer  to  the  analogous  problem   (Art. 

205): 

"  Find  a  number  such  that  twice  its  square  diminished  by 
three  times  the  number  may  be  Go." 

2.  A  farmer  bought  some  sheep  for  $  72,  and  found  that  if 
he  had  bought  6  more  for  the  same  money,  he  would  have  paid 
$  1  less  for  each.     How  many  sheep  did  he  buy  ? 

Let  x  =  the  number  of  sheep  bought. 

72 
Then  —  =  the  price  paid  for  one, 


x 

72 
and =  the  price  paid,  if  6  more. 

JO     \~   O 

72         72 

By  the  conditions,      —  = -  +  1 

J  '       x       x  +  6 

Whence,  x  =  18  or  —  24. 

Here  the  negative  result  is  not  admissible  as  a  solution  of 
the  problem  in  its  present  form;  the  number  of  sheep,  there- 
fore, was  18. 

If,  in  the  given  problem,  "6  more  "  be  changed  to  "6  fewer" 
and  "$1  less"  to  "$1  more,"  24  will  be  the  true  answer. 

Hence,  we  infer  that 

A  negative  result,  obtained  as  one  of  the  answers  to  a  prob- 
lem, is  sometimes  the  answer  to  another  analogous  problem, 
formed  by  attributing  to  the  unknown  quantity  a  quality 
directly  opposite  to  that  which  has  been  attributed  to  it. 


DISCUSSION   OF  PROBLEMS.  259 

INTERPRETATION   OF  IMAGINARY   RESULTS. 

335.  It  lias  been  shown  (Art.  324)  under  what  circum- 
stances a  quadratic  equation  will  be  in  form  to  produce  imagi- 
nary roots.     It  is  now  proposed  to  interpret  such  results. 

Let  it  be  required  to  divide  10  into  two  such  parts  that  their 
product  shall  be  26. 

Let  x  =  one  of  the  parts. 

Then  10  —  x  =  the  other. 

By  the  conditions,  x  (10  —  x)  =  26 


Whence,  x  =  5  ±  y/— 1. 

Thus,  we  obtain  an  imaginary  result.  We  therefore  con- 
clude that  the  problem  cannot  be  solved  numerically  ;  in  fact, 
if  we  call  one  of  the  parts  5  +  y,  the  other  must  be  5  —  y,  and 
their  product  will  be  25  —  y'2.  which,  so  long  as  y  is  numerical, 
is  less  than  25.  But  we  are  required  to  find  two  numbers 
whose  sum  is  10  and  product  26 ;  there  are,  then,  no  such 
numbers. 

Had  it  been  required  to  find  two  expressions,  whose  sum  is 

10    and   product    25.    the    answer    5  +  \/ —  1    and    5  —  \j—  1 
would  have  satisfied  the  conditions. 

The  given  problem,  however,  expresses  conditions  incom- 
patible with  each  other,  and,  consequently,  is  impossible. 
Hence, 

Imaginary  results  indicate  that  the  problem  is  impossible. 

PROBLEM   OF   THE   LIGHTS. 

336.  The  principles  of  interpretation  will  be  further  illus- 
trated in  the  discussion  of  the  following  general  problem. 

Find  upon  the  line  which  joins  two  lights,  A  and  B,  the 
point  which  is  equally  illuminated  by  them  :  admitting  that 
the  intensity  of  a  light,  at  a  given  distance,  is  equal   to  its 


2G0  ALGEBRA. 

intensity  at  the  distance  1,  divided  by  the  square  of  the  given 
distance. 

C"  A  C  B  C 

— I 1 J 1 1 — 

Assume  A  as  the  origin  of  distances,  and  regard  all  dis- 
tances estimated  to  the  right  as  positive. 

Let  a  denote  the  intensity  of  the  light  A,  at  the  distance  1 ; 
b  the  intensity  of  the  light  B,  at  the  distance  1 ;  and  c  the 
distance  A  B,  between  the  two  lights. 

Suppose  C  the  point  of  equal  illumination,  and  let  x  repre- 
sent the  distance  from  it  to  A,  or  the  distance  AC.  Then, 
c  —  x  will  represent  the  distance  B  C. 

By  the  conditions  of  the  problem,  since  the  intensity  of  the 

light  A,  at  the  distance  1,  is  a,  at  the  distance  x  it  is  — =  :  and 

X' 

since  the  intensity  of  the  light  B,  at  the  distance  1,  is  b,  at  the 

distance  c  —  x  it  is  -. ^ .     But,  by  supposition,  at  C  these 

intensities  are  equal ;  hence, 
a  b 


Whence, 


x2    '  (c  -  xf  ' 
c—x 


~  (c- 

x)2 

b 

or 

X2 

a 

_+v* 

x  \l  a 

From  this  equation  we  obtain  as  the  values  of  x : 

c\J  a  i      \J  a 


x 


or, 


\Ja  +  ^b      w\\ja  +  ^by 

c\J  a  /      sj  a       \ 

\J  a  —  \J  b         \\J a  —  \J b)' 


Since  both  a  and  b  are  positive,  the  two  values  of  x  are  both 
real.     Hence, 

There  are  two  points  of  equal  illumination  on  the  line  of  the 
lights. 


DISCUSSION   OF  PROBLEMS.  261 

Since  there  are  two  lights,  c  must  always  he  greater  than  0 ; 
consequently  neither  a,  b,  nor  c  can  he  0.  The  problem,  then, 
admits  properly  of  only  these  three  different  suppositions : 

1.    a  >  b.  2.    a  <  b.  3.    a=b. 

We  shall  now  discuss  the  values  of  x  under  each  of  these 
suppositions. 

1.    a  >  b. 

In  this  case,  the  first  value  of  x  is  less   than  c ;   because 

- being  a  proper  fraction,  is  less  than  1.      This  value 

y/  a  +  \J  b 

c 

of  x  is  also  greater  than  - ;  because,  the  denominator  being 

less  than  twice  the  numerator,  as  b  is  less  than  a,  the  fraction 
is  greater  than  \.  Hence,  the  first  point  of  equal  illumination 
is  at  C,  between  the  two  lights,  but  nearer  the  lesser  one. 

The  second  value  of  x  is  greater  than  c :  because  —A- —  , 

ya  —  \J  b 

being  an  improper  fraction,  is  greater  than  1.  Hence,  the 
second  point  is  at  C,  in  the  prolongation  of  the  line  A  B,  be- 
yond the  lesser  light. 

These  results  agree  with  the  supposition.  For,  if  a  is  greater 
than  b,  then  B  evidently  is  the  lesser  light.  Hence,  both  points 
of  equal  illumination  will  be  nearer  B  than  A ;  and  since  the 
two  lights  emit  rays  in  all  directions,  one  of  the  points  must 
he  in  the  prolongation  of  A  B  beyond  both  lights. 

2.    a  <  b. 

In  this  case,  the  first  value  of  x  is  positive.     It  is  also  less 

than   C-\   because     .    ^  C '   . ,  ,  having  the  denominator  greater 
2  v  a  +  y  b 

than  twice  the  numerator,  b  being  greater  than  a,  is  less  than 

h-     Hence,  the  first  point  of  equal  illumination  is  between  the 

lights,  but  nearer  A,  the  lesser  light, 

The  second  value  of  x  is  negative,  because  the  denominator 

y/  a  —  y1  b  is  negative  ;  which  must  he  interpreted  as  measur- 


262  ALGEBKA. 

ing  distance  from  A  towards  the  left  (Art.  205).  Hence,  the 
second  point  of  equal  illumination  is  at  C",  in  the  prolongation 
of  the  line,  at  the  left  of  the  lesser  light,  A. 

These  results  correspond  with  the  supposition;  the  case 
being  the  same  as  the  preceding  one,  except  that  A  is  now  the 
lesser  light. 

3.    a  =  b. 
In  this  case,  the  first  value  of  x  is  positive,  and  equal  to  ^  . 

Li 

Hence,  the  first  point  of  equal  illumination  is  midway  be- 
tween the  two  lights. 

The  second  value  of  x  is  not  finite:  because  -; — p- ,  if 

V  «  —  V  b 

I 

a  =  b,  reduces  to  ~-  =  cc   (Art.  210),  which  indicates  that  no 

finite  value  can  be  assigned  to  x.  Hence,  there  is  no  second 
point  of  equal  illumination  in  the  line  A  B,  or  its  prolongation. 

These  results  agree  with  the  supposition.  For,  since  the 
lights  are  of  equal  intensity,  a  point  of  equal  illumination  will 
obviously  be  midway  between  them ;  and  it  is  evident  that 
there  can  be  no  other  like  point  in  their  line. 

The  preceding  discussion  illustrates  the  precision  with  which 
algebraic  processes  will  conform  to  every  allowable  interpreta- 
tion of  the  enunciation  of  a  problem. 


XXXI.  — RATIO  AND  PROPORTION. 

337.  The  Ratio  of  one  quantity  to  another  of  the  same 
kind  is  the  quotient  arising  from  dividing  the  first  quantity 
by  the  second  (Art.  181). 

Thus,  the  ratio  of  a  to  b  is  - ,  or  a  :  b. 

b 


EATIO   AND   PROPORTION.  263 

338.  The  Terms  of  a  ratio  are  the  two  quantities  required 
to  form  it.  Of  these,  the  first  is  called  the  antecedent,  and  the 
second  the  consequent. 

Thus,  in  the  ratio  a  :  b,  a  and  b  are  the  terms,  a  the  ante- 
cedent, and  b  the  consequent. 

339.  A  Proportion  is  an  equality  of  ratios  (Art.  181). 
Thus,  if  the  ratios  a  :  b  and  c  :  d  are  equal,  they  form  a  pro- 
portion, which  may  he  written 

a  :  b  =  c  :  d,  or  a  :  b  : :  c  :  d. 

340.  The  Terms  of  a  proportion  are  the  four  terms  of  its 
two  ratios.  The  first  and  third  terms  are  called  the  antece- 
dents ;  the  second  and  fourth,  the  consequents;  the  first  and 
last,  the  extremes  ;  the  second  and  third,  the  means  ;  and  the 
terms  of  each  ratio  constitute  a  couplet. 

Thus,  in  a  :  b  =  c  :  d,  a  and  c  are  antecedents ;  b  and  d,  con- 
sequents;  a  and  d,  extremes;  b  and  c,  means;  a  and  b,  the 
first  couplet ;  and  c  and  d,  the  second  couplet. 

341.  A  Proportional  is  any  one  of  the  terms  of  a  propor- 
tion ;  a  Mean  Proportional  between  two  quantities  is  either 
of  the  two  means,  when  they  are  equal ;  a  Third  Proportional 
to  two  quantities  is  the  fourth  term  of  a  proportion,  in  which 
the  first  term  is  the  first  of  the  quantities,  and  the  second  and 
third  terms  each  equal  to  the  second  quantity ;  a  Fourth  Pro- 
portional to  three  quantities  is  the  fourth  term  of  a  proportion 
whose  first,  second,  and  third  terms  are  the  three  quantities 
taken  in  their  order. 

Tims,  in  a  :  b  =  b  :  e,  b  is  a  mean  proportional  between  a 
and  c  ;  and  c  is  a  third  proportional  to  a  and  b.  In  a  :  b  —  c  :  d, 
d  is  a  fourth  proportional  to  a,  b,  and  c. 

342.  A  Continued  Proportion  is  one  in  which  each  conse- 
quent is  the  same  as  the  next  antecedent ;  as, 

a  :  b  =  b  :  c  =  c  :  d  =■  d  :  e. 


264  ALGEBRA. 

PROPERTIES   OF  PROPORTIONS. 

343.      Wlien  four  quantities  are  in  proportion-,  the  prodwt 
of  the  extremes  is  equal  to  the  product  of  the  means. 

Let  the  proportion  be 

a  :  b  =  c  :  d. 

This  may  he  written  (Art.  337), 

a      c 
b=d 

Whence,  ad  =  b  c. 

Hence,  if  three  quantities  be  in  continued  proportion,  the 
product  of  the  extremes  is  equal  to  the  square  of  the  means. 

Thus,  if 

a  :  b  =  b  :  c 

then,  a  c  =  b2. 

By  this  theorem,  if  three  terms  of  a  proportion  are  given, 
the  fourth  may  be  found.     Thus,  if 


then, 
Whence, 

344.  If  the  product  of  two  quantities  be  equal  to  the  prod- 
uct of  two  others,  one  pair  of  them  may  be  made  the  extremes, 
and  the  other  pair  the  means,  of  a  proportion. 

Thus,  if 

a  d  =  b  c 

ad       be         a      c 

Dividing  bv  b  d,  -=— =■  =  rr— ;  ,  or  T  =  -= 

&    J  bd       bd'       b       d 

Whence  (Art.  337),  a:b  =  c:d. 


a  :  b 

=  c  :  x 

a  x 

=  bc 

x 

_bc 

a 

RATIO  AND  PROPORTION.  265 

In  a  similar  manner,  we  might  derive  from  the  equation 
a  d  =  b  c,  the  following  proportions : 

a  :  c  =  b  :  c 1, 

b  :  d  =  a  :  c, 

c  :  d  =  a  :  b, 

d  :  b  =  c  :  a,  etc. 

345.  If  four  quantities  are  in  proportion,  they  will  be  in 
proportion  by  Alternation;  that  is,  the  antecedents  will 
have  to  each  other  the  same  ratio  as  the  consequents. 

Thus,  if  a  :  b  =  c  :  d 

then  (Art.  343),  ad  =  bc 

Whence  (Art.  344),  a  :  c  =  b  :  d. 

346.  If  four  quantities  are  in  proportion,  they  will  be  in 
proportion  by  Inversion;  that  is,  the  second  term  will  be  to 
the  first,  as  the  fourth  is  to  the  third. 

Thus,  if  a  :  b  —  c  :  d 

then,  ad  — be 

Whence,  b  :  a  =  d  :  c. 

347.  If  four  quantities  are  in  proportion,  they  will  be  in 
proportion  by  Composition;  that  is,  the  sum  of  the  first  two 
terms  ivill  be  to  the  first  term,  as  the  sum  of  the  last  two  terms 
is  to  the  third  term. 

Thus,  if  a  :  b  —  c  :  d 

then,  ad  =  b  c 

Adding  hoth  members  to  a  c, 

ac  +  ad  =  ac  +  bc,  or  a  (c '  +  d)  =  c  {a  +  b) 

Whence,  a  +  b:  a  =  c  +  d:  c  (Art.  344). 

Similarly,  we  may  show  that 

a  +  b  :  b  —  c  +  d  :  d. 


266  ALGEBRA. 

348.  If  four  quantities  are  in  proportion,  they  will  be  in 
proportion  by  Division;  that  is,,  the  difference  of  the  first  two 
terms  will  be  to  the  first  term,  as  the  difference  of  the  last  two 
terms  is  to  the  third  term. 

Thus,  if  a  :  b  =  c  :  d 

then,  a  d  =  b  c 

Subtracting  both  members  from  a  c, 

ac  —  ad  =  ac  —  be,  or  a  (c  —  d)=c  (a  —  b) 
Whence,  a  —  b  :  a  =  c  —  d  :  c. 

Similarly,  we  may  prove  that 

a  —  b  :  b  =  c  —  d  :  d. 

349.  If  four  quantities  are  in  proportion,  they  will  be  in 
proportion  by  Composition  and  Division;  that  is,  the  sum 
of  the  first  two  terms  will  be  to  their  difference,  as  the  sum  of 
the  last  two  terms  is  to  their  difference. 


(1) 
(2) 


Thus,  if 

a  :  b  —  c  :  d 

by  Art.  347, 

a  +  b      c  +  d 
a             e 

and,  by  Art.  348, 

a  —  b      c  —  d 
a             c 

Dividing  (1)  by  (2), 

a  +  b      c  +  d 
a  —  b      c  —  d 

or,                            a  -f-  b 

: a  —  b  =  c  +  d : c 

d. 

350.     Quantifies  which  are  proportional  to  the  same  quan- 
tities, are  propjortional  to  each  other. 

Thus,  if  a  :b  =  e:f 

and  c  :  d  =  e  :f 

.  ae.ee 

then,  r  =  3   and   -7  =  ^ 

Therefore,  -=- 

b      d 

Whence,  a:b  =  c:d. 


RATIO  AND   PROPORTION.  267 

351.  If  any  number  of  quantities  are  proportional,  any 

antecedent  is  to  its  consequent,  as  the  sum  of  all  the  antece- 
dents is  to  the  sum  of  all  the  consequents. 

Thus,  if 

a  :  b  =  c  :  d  =  e  :f 

then  (Art.  343),  ad  =  b  c 

and  af-=be 

also,  a  b  =  «  b 

Adding,  a  (b  +  d  +/)  =  b  (a  +  c  +  e) 

Whence  (Art.  344),  a  :b  =a  +  c  +  e  :  b  +  d +f 

352.  If  four  quantities  are  in  proportion,  if  the  first  and 
second  be  multiplied  or  divided  by  any  quantity,  as  also  the 
third  and  fourth,  the  resulting  quantities  will  be  in  proportion* 

Thus,  if 

a  :  b  =  c  :  d 


then, 

a      c 
b=d 

Therefore, 

ma      nc 
m  b       n  d 

Whence, 

m  a  :  mb  =  n  c  :  n  d. 

In  a  similar 

manner  we  could  prove 

a       b        c     d 

m  '  m       n'  n' 

Either  m  or  n  may  be  made  equal  to  unity.  That  is,  either 
couplet  may  be  multiplied  or  divided,  without  multiplying  or 
dividing  the  other. 

353.  If  four  quantities  are  in  proportion,  and  the  first  ami 
third  be  multiplied  or  divided  by  any  quantity,  as  also  the 
second  and  fourth,  the  resulting  quantities  will  be  in  jiro- 
portion. 


268  ALGEBRA. 


Thus,  if  a  :  b  =  c  :  d 


then, 


Therefore, 


a c 

b=d 

m  a      m  c 


nb       n  d 

Whence,  m  a  :  nb  =  m  e  :  nd. 

In  a  similar  manner  we  could  prove 

a     b  _     c      d 

m     n      m  '  n 

Either  m  or  n  may  he  made  equal  to  unity. 

354.     If  there  be  two  sets  of  proportional  quantities,  the 
products  of  the  corresponding  terms  will  be  in  proportion. 

Thus,  if  a  :b  =  c  :  d 

and  e:f=g:h 

a      c         ..    e       q 
then,  -  =  -    and    -=- 


Therefore, 


b      d  f     h 

ae      eg 


bf      dh 
Whence,  ae  :bf=  c  g  :  d  h. 

355.  If  four  quantities  are  in  proportion,  like  poioers  or 
like  roots  of  these  quantities  will  be  in  proportion. 

Thus,  if  a  :b  =  c  :d 

then,  r  =  --,',  therefore,  —  =  — 

b      d  '  bn       dn 

Whence,  an  :bn  =  cn  :  dn. 

In  a  similar  manner  we  could  prove 

y/  a  :  y/  b  =  y/  c  :  y  d. 

356.  If  three  quantities  are  in  continued  proportion,  the 
first  is  to  the  third,  as  the  square  of  the  first  is  f>>  the  square 
of  the  second. 


RATIO   AND   PROPORTION.  269 


Thus,  if  a:b  =  b  :  c 

a      b 


then, 


a 


b      c 
a2       aba 


Multiplying  by -,  y=l*c      c 

Whence,  a:  c  —  a~ :  b~. 

In  a  similar  manner  we  could  prove  that  if 

a  :  b  =  b  :  c  =  c  :  d,  then  a\d=-az:  b3. 

Note.     The  ratio  a2  :  b2  is  called  the  duplicate  ratio,  and  the  ratio  a?  :  b3 
the  triplicate  ratio,  of  a  :  b. 

PROBLEMS. 

357.     1.    The  last  three  terms  of  a  proportion  being  18,  6, 
and  27,  what  is  the  first  term  ? 

2.  The  first,  third,  and  fourth  terms  of  a  proportion  being 
4,  20,  and  55,  respectively,  what  is  the  second  term  ? 

3.  Find  a  fourth  proportional  to  \,  \,  and  \. 

4.  Find  a  third  proportional  to  5  and  3. 

5.  Find  a  mean  proportional  between  2  and  8. 

6.  Find  a  mean  proportional  between  6  and  24. 

7.  Find  a  mean  proportional  between  49  and  4. 

8.  Find  two  numbers  in  tbe  ratio  of  2\  to  2,  such  that  when 
each  is  diminished  by  5,  they  shall  be  in  the  ratio  of  lj  to  1. 

9.  Divide  50  into  two  such  parts  that  the  greater  increased 
by  3,  shall  be  to  the  less  diminished  by  3,  as  3  to  2. 

10.  Divide  27  into  two  such  parts  that  their  product  shall 
be  to  the  sum  of  their  squares  as  20  to  41. 

11.  There  are  two  numbers  which  are  to  each  other  as  4  to 
9,  and  12  is  a  mean  proportional  between  them.  What  are 
the  numbers  ? 


270  ALGEBRA. 

12.  The  sum  of  two  numbers  is  to  their  difference  as  10  to 
3,  and  their  product  is  364.     What  are  the  numbers-  ? 

13.  Find  two  numbers  such  that  if  3  be  added  to  each,  they 
will  be  in  the  ratio  of  4  to  3  ;  and  if  8  be  subtracted  from  each, 
they  will  be  in  the  ratio  of  9  to  4. 

14.  There  are  two  numbers  whose  product  is  96,  and  the 
difference  of  their  cubes  is  to  the  cube  of  their  difference  as  19 
to  1.     What  are  the  numbers  ? 

15.  Each  of  two  vessels  contains  a  mixture  of  wine  and 
water ;  a  mixture  consisting  of  equal  measures  from  the  two 
vessels,  contains  as  much  wine  as  water;  and  another  mixture 
consisting  of  four  measures  from  the  first  vessel  and  one  from 
the  second,  is  composed  of  wine  and  water  in  the  ratio  of  2  to 
3.     Find  the  ratio  of  wine  to  water  in  each  vessel. 

16.  If  the  increase  in  the  number  of  male  and  female 
criminals  be  1.8  per  cent,  while  the  decrease  in  the  number  of 
males  alone  is  4.6  per  cent,  and  the  increase  in  the  number  of 
females  alone  is  9.8  per  cent,  compare  the  number  of  male  and 
female  criminals,  respectively,  at  the  first  time  mentioned. 

17.  A  railway  passenger  observes  that  a  train  passes  him, 
moving  in  the  opposite  direction,  in  2  seconds ;  whereas,  if  it 
had  been  moving  in  the  same  direction  with  him,  it  would 
have  passed  him  in  30  seconds.  Compare  the  rates  of  the  two 
trains. 


XXXII.  —  VARIATION. 

358.  Variation,  or  general  proportion,  is  an  abridged 
method  of  expressing  common  proportion. 

Thus,  if  A  and  B  be  two  sums  of  money  loaned  for  equal 
times,  at  the  same  rate  of  interest,  then 

A  :  B  =  A's  interest  :  B's  interest 


VARIATION.  271 

or,  in  an  abridged  form,  by  expressing  only  two  terms,  the  in- 
terest varies  as  the  principal ;  thus  (Art.  23), 

The  interest  oc  the  principal. 

359.  One  quantity  varies  directly  as  another,  when  the 
two  increase  or  decrease  together  in  the  same  ratio. 

Sometimes,  for  the  sake  of  brevity,  we  say  simply  one  quan- 
tity varies  as  another,  omitting  the  word  "  directly/' 

Thus,  if  a  man  works  for  a  certain  sum  per  day,  the  amount 
of  his  wages  varies  as  the  number  of  days  during  which  he 
works.  For,  as  the  number  of  days  increases  or  decreases,  the 
amount  of  his  wages  will  increase  or  decrease,  and  in  the  same 

ratio. 

360.  One  quantity  varies  inversely  as  another,  when  the 
first  varies  as  the  reciprocal  of  the  second. 

Thus,  if  a  courier  has  a  fixed  route,  the  time  in  which  he 
will  pass  over  it  varies  inversely  as  his  speed.  That  is,  if  he 
double  his  speed,  he  will  go  in  half  the  time;  and  so  on. 

361.  One  quantity  varies  as  two  others  jointly,  when  it 
has  a  constant  ratio  to  the  product  of  the  other  two. 

Thus,  the  wages  of  a  workman  will  vary  as  the  number  -of 
days  he  has  worked,  and  the  wages  per  day,  jointly. 

362.  One  quantity  varies  directly  as  a  second  and  inversely 
as  a  third,  when  it  varies  jointly  as  the  second  and  the  recip- 
rocal of  the  third. 

Thus,  in  physics,  the  attraction  of  a  planetary  body  varies 
directly  as  the  quantity  of  matter,  and  inversely  as  the  square 
of  the  distance. 

363.  If  A  varies  as  B,  then  A  is  equal  to  B  multiplied  by 
some  constant  quantity. 

Let  a  and  b  denote  one  pair  of  corresponding  values  of  two 
quantities,  and  A  and  B  any  other  pair.     Then,  by  Art.  358, 


272  ALGEBRA. 

A:a  =  B:b 

Whence  (Art.  343),  Ab  =  a  B,  or  A=^B 

•    a 
Denoting  the  constant  ratio  -  by  m, 

A  =  m  B. 

Hence,  also,  when  any  quantity  varies  as  another,  if  any 
two  pairs  of  values  of  the  quantities  be  taken,  the  four  will  be 
proportional. 

For,  if  A  oc  B,  and.  A'  and  B'  be  any  pair  of  values  of  A  and 
B,  and  A"  and  B"  any  other  pair,  by  Art.  363, 

A'  =  m  B',  and  A"  =  m  B" 


a 


A1  A 

Whence,  -~  =  m,  and  —  - 


Therefore, 


A" 


B1      B" 

or  (Art.  337),  A':B'  =  A"  :  B 


'ii 


364.     The  terms  used  in  Variation   may  now  be  distin- 
guished as  follows : 

1.  If  A  =  m  B,  A  varies  directly  as  B. 

Ml 

2.  If  A  =  — ,  A  varies  inversely  as  B. 

3.  If  A  =  m  B  C,  A  varies  jointly  as  B  and  C. 

4.  If  A  —  ,  A  varies  directly  as  B,  and  inversely  as  C. 

Problems  in  variation,  in  general,  arc  readily  solved  by  con- 
verting the  variation  into  an  equation,  by  the  aid  of  Art.  364. 


VAKIATION.  273 

EXAMPLES. 

365.  1.  Given  that  y  oc  x,  and  when  x  =  2,  y  =  10.  Re- 
quired the  value  of  y  in  terms  of  x. 

If  y  oc  x,  by  Art.  364,  y  =  mx 

Substituting  x  =  2  and  y  =  10,  10  =  2  m,  whence  m  =  5. 
Hence,  the  required  value  is  y  =  5  a?. 

2.  Given  that  7/  oc  »■,  and  that  v/  =  2  when  x  =  l.  What 
will  he  the  value  of  y  when  x  =  2  ? 

3.  If  y  oc  £,  and  y  =  21  when  £  =  3,  find  the  value  of  y  in 
terms  of  z. 

4.  If  x  varies  inversely  as  y,  and  x  =  4  when  y  =  2,  what  is 
the  value  of  cc  when  y  =  6? 

5.  Given  that  z  varies  jointly  as  x  and  y,  and  that  z  =  1 
when  x  =  1  and  ?/  =  1.  Find  the  value  of  z  when  x  =  2  and 
y  =  2. 

6.  If  ?/  equals  the  sum  of  two  quantities,  of  which  one  is 
constant,  and  the  other  varies  as  x  y  ;  and  when  x  =  2,  y  =  —  2  J-, 
hut  when  x  =  —  2,  y  =  1 ;  express  y  in  terms  of  x. 

7.  Two  circular  plates  of  gold,  each  an  inch  thick,  the  diam- 
eters of  which  are  6  inches  and  8  inches,  respectively,  are 
melted  and  formed  into  a  single  circular  plate  one  inch  thick- 
rind  its  diameter,  having  given  that  the  area  of  a  circle  varies 
as  the  square  of  its  diameter. 

8.  Given  that  the  illumination  from  a  source  of  light  varies 
inversely  as  the  square  of  the  distance  ;  how  much  farther  from 
a  candle  must  a  book,  which  is  now  3  inches  off,  be  removed, 
so  as  to  receive  just  half  as  much  light  ? 

'  9.  A  locomotive  engine  without  a  train  can  go  24  miles  an 
hour,  and  its  speed  is  diminished  by  a  quantity  which  varies 
as  the  square  root  of  the  number  of  cars  attached.  With  four 
cars  its  speed  is  20  miles  an  hour.  Find  the  greatest  number 
of  cars  which  the  engine  can  move. 


274  ALGEBEA. 

XXXIII.  —  ARITHMETICAL  PROGRESSION. 

366.  An  Arithmetical  Progression  is  a  series  of  quanti- 
ties, in  which  each  term  is  derived  from  the  preceding  term 
by  adding  a  constant  quantity,  called  the  com raon  difference. 

367.  When  the  series  is  increasing,  as,  for  example, 

1,3,5,7,9,11, 

each  term  is  derived  from  the  preceding  term  by  adding  a 
positive  quantity;  consequently  the  common  difference  is 
positive. 

When  the  series  is  decreasing,  as,  for  example, 

19,  17,15,13,11,9,  

each  term  is  derived  from  the  preceding  term  by  adding  a 
negative  quantity;  consequently  the  common  difference  is 
negative. 

368.  Given  the  first  term,  a,  the  common  difference,  d,  and 
the  number  of  terms,  n,  to  find  the  last  term,  I. 

The  progression  will  be 

a,  a  +  d,  a  +  2  d,  a  +  3  d, 

We  observe  that  these  terms  differ  only  in  the  coefficient  of 
d,  which  is  1  in  the  second  term,  2  in  the  third  term,  3  in  the 
fourth  term,  etc.  Consequently  in  the  rath  term,  the  coefficient 
of  d  will  be  n  —  1.  Hence,  the  rath  term  of  the  series,  or  the 
last  term,  as  the  number  of  the  terms  is  n,  will  be 

l  =  a+  (n-l)d  (1) 

369.  Given  the  first  term,  a,  the  last  term,  1,  and  the  num- 
ber of  terms,  n,  to  find  the  sum  of  the  series,  S. 

S=a+  (a  +  d)  +  (a  +  2d)  + +  (l-2d)  +  (l  —  d)  +  l 

Writing  the  terms  of  the  second  member  in  the  reverse 
order, 

S=l+  (l-d)  +  (I-  2  d)  + +  {a  +  2  d)  +  (a  + d)  +  a 


ARITHMETICAL  PROGRESSION.  275 

Adding  these  equations,  term  by  term,  we  have 
2S=(a+l)  +  (a  +  l)  +  (a  +  l)+ +  (a+t)+(a+t)  +  (a+l) 

In  this  result,  (a  +  I)  is  taken  as  many  times  as  there  are 
terms,  or  n  times ;  hence 

2S=n(a+l),ovS=7i(a+I)  (2) 

Using  the  value  of  I  given  in  (1),  Art,  3G8,  this  may  he 
written 

S=%[2a  +  (n-l)d] 

370.     1.    In  the  series  5,  8/11, to  18  terms,  find  the 

last  term  and  the  sum  of  the  series. 

Here  a  =  5,  n  =  18  ;  the  common  difference  is  always  found 
by  subtracting  the  first  term  from  the  second;  hence 
d  =  8-5  =  3. 

Substituting  these  values  in  (1)  and  (2),  we  have 

I  =  5  +  (18  - 1)  3  =  5  +  17  x  3  =  5  +  51  =  56. 

1 S 
S  =  -^  (5  +  56)  =  9  x  61  =  549. 

2.    In  the  series  2,-1,-4, to  27  terms,  find  the  last 

term  and  the  sum  of  the  series. 

Here  a  =  2,  n  =  27,  d  =  the  second  term  minus  the  first 
=  —  1  —  2  =  —  3.  Substituting  these  values  in  (1)  and  (2), 
we  have 

1  =  2  +  (27-1)  (-3)  =  2  +  26  (-3)  =2-  78  =  -76. 

S=^-(2-76)=^-(-74)  =  27(-37)  =  -999. 

« 

EXAMPLES. 

Find  the  last  term  and  the  sum  of  the  series  in  the  fol- 
lowing : 


276  ALGEBRA. 

3.  1,  6,  11,  to  15  terms. 

4.  7,  3,  —  1,  to  20  terms. 

5.  —  9,  —  6,-3,  to  23  terms. 

6.  -5,-10,-15,  to  29  terms. 

^    /    o    o  . 

o>  7>  k>   to  16  terms. 

8.  ■=,  T-p ,   to  19  terms. 

5     15 

1      5 

9.  - ,  — ,   to  22  terms. 

2    1 

10.  —  - ,  - ,   to  14  terms. 

o     o 

5 

11.  — 3,  — -,   to  17  terms. 

113 

12.  j  j  9  j  j  3   to  35  terms. 

371.  Formulae  (1)  and  (2)  constitute  two  independent 
equations,  together  containing  all  the  five  elements  of  an 
arithmetical  progression  ;  hence,  when  any  three  of  the  five 
elements  are  given,  we  may  readily  deduce  the  values  of  the 
other  two,  as  by  substituting  the  three  known  values  we  shall 
have  two  equations  with  only  two  unknown  quantities,  which 
may  be  solved  by  methods  previously  given. 

1.  The  first  term  of  an  arithmetical  progression  is  3,  the 
number  of  terms  20,  and  the  sum  of  the  terms  440.  Find  the 
last  term  and  the  common  difference. 

Here  a  =  3,  n  =  20,  £=440;  substituting  in  (1)  and  (2), 
we  have 

1  =  3 +  19  d 

-  440  =  10  (3  +  I),  or  44  =  3  +  I 

From  the  second  equation,  I  =  41  ;    substitute  in  the  first, 

41  =  3  +  19  d ;  19  d  =  38 ;  d  =  2. 


ARITHMETICAL   PROGRESSION.  277 

2.  Given  d  =  —  3,  I  =  —  39,  S  =  —  264 ;  find  a  and  n. 
Substituting  the  given  quantities  in  (1)  and  (2), 

-  39  =  a  +  (n  -  1)  (-  3),  or  3  n  -  a  =  42 

-  264  =  -  (a  -  39),  or  a  n  —  39  n  =  -  528 

From  the  first  of  these  equations,  a  —  3  ?i  —  42  ;  substitute 
in  the  second, 

(3  n  -  42)  n  -  39  »  =  -  528,  or  ?r  -  27  n  =  -  176 

Whence,  n  =  - ~ =  ^—^ —  =  16  or  11 

Substituting  in  the  equation  a  =  3n  —  42, 

When  ii  —  16,  a  =  6 

n  =  11,  a.  =  —  9,  Ans. 

The  signification  of  the  two  answers  is  as  follows : 
If  n  =  16,  and  a  =  6,  the  series  will  be 

6,  3,  0,  -  3,  -  6,  -  9,  - 12,  -  15,  - 18,  -  21,  -  24,  -  27, 

-  30,  -  33,  -  36,  -  39. 

If  n  =  11,  and  a  =  —  9,  the  series  will  be 
_  o,  - 12,  - 15,  - 18,  -  21,  -  24,  -  27,  -  30,  -  33,  -  36,  -  39. 
In  either  of  which  the  last  term  is  —  39  and  the  sum  —  264. 

113 

3.  Given  a  =  ^ ,  d  —  —Tx,  S  =  —  ^;  find  I  and  n. 

Substituting  the  given  quantities  in  (1)  and  (2),  we  have 
J  =  |+(»_l)(-l),or  12l  +  n*=5 


S_nfl 

'2~2 


(=  +  l) ,  ox  n  +  3 1  n  =  —  9 


278  ALGEBRA. 

From,  the   first   of   these,  n  =  5  —  12  I ;  substitute   in   the 
second, 

5  - 12 1  +  3 I  (5  - 12 I)  =  -  9,  or  36  V"  -  3 1  =  14 

,      3  ±y/  9  +  2016     3  +  45     2  7 

Whence,      J  = ^ =    70     =  3  °r  ~~  12 


Substituting  in  the  equation  re  =  5  — 12  £, 


9 


"When  I  =  o  ,  n  —  —  3 


o 


1  =  —  —  ,  n  =  12,  Ans. 

The  first  answer  is  inapplicable,  as  a  negative  number  of 
terms  has  no  meaning.     Hence  the  only  answer  is, 

7 
l  =  —  j2>  ft  =  12. 


Note.     A  negative  or  fractional  value  of  n  is  always  inapplicable,  and 
should  be  neglected,  together  with  all  other  values  dependent  on  it. 


EXAMPLES. 

4.  Given  d  =  4,  1  =  75,  re  =  19  ;  find  a  and  >Sf. 

165 

5.  Given  cZ  =  —  1,  re  =  15,  $  = — -  ;  find  a  and  Z. 

-j 

2 

6.  Given  a  =  —  -,  re  =  18,  £  =  5 ;  find  d  and  & 

o 

3 

7.  Given  a  =  —  — ,  re  =  7,  $  =  —  7  ;  find  d  and  Z. 

8.  Given  I  =  —  31,  re  =  13,  S=  -  169  ;  find  a  and  tf. 

9.  Given  a  =  3,  I  =  42-f ,  d  =  2£ ;  find  re  and  S. 

10.  Given  a  =  7,  l  =  —  73,  #=—  363  ;  find  re  and  c?. 

n     n-  15  5  2625 

11.  Given  a  =  — ,  d  =  7:,  o=  ;  find  re  and  Z. 

£  Ji  Ji 


ARITHMETICAL   PROGRESSION.  279 

12.  Given  I  =  -  47,  d  =  -  1,  8=  -  1118  ;  find  a  and  n. 

13.  Given  d  =  —  3,  S  —  —  328,  a  =  2;  find  Z  and  n. 

372.  2b  insert  any  number  of  arithmetical  means  between, 
two  given  terms. 

1.  Insert  5  arithmetical  means  between  3  and  —  5. 

This  may  he  performed  in  the  same  manner  as  the  examples 
in  the  previous  article ;  we  have  given  the  first  term  a  =  3 ; 
the  last  term  I  =  —  5  ;  the  number  of  terms  n  =  7  ;  as  there 
are  5  means  and  two  extremes,  or  in  all  7  terms.  Substituting 
in  (1),  Art.  368,  we  have 

4 

—  5  =  3  +  6  d ;  or,  6  d  =  —  8  ;  whence,  d  =  —  -= . 

o 

4 
Hence  the  terms  are  obtained  by  subtracting  -  from  3  for 

4 

the  first,  -  from  that  result  for  the  second,  and  so  on ;  or, 

35    1        1  -l    -11    -5  Ans 
«5j  o  i  o  j  —  Lt  —  o  j o"  >       °>  sins. 

EXAMPLES. 

2.  Insert  5  arithmetical  means  between  2  and  4. 

3.  Insert  7  arithmetical  means  between  3  and  —  1. 

4.  Insert  4  arithmetical  means  between  —  1  and  —  6. 

5.  Insert  6  arithmetical  means  between  —  8  and  —  4. 

6.  Insert  4  arithmetical  means  between  —  2  and  6. 

7.  Insert  m  arithmetical  means  between  a  and  b. 

PROBLEMS. 

373.  1.  The  6th  term  of  an  arithmetical  progression  is  19, 
and  the  14th  term  is  67.     Find  the  first  term. 

By  Art.  368,  the  6th  term  is  a  +  5  d,  and  the  14th  term  is 
a  +  13  d.     Hence, 


280  ALGEBRA. 

a+    5d  =  19 
a  ±13  d  =  67 


Whence,  8  d  =  48,  or  d  =  6 

Therefore,  a  =  —  11,  Ans. 

2.  Find  four  quantities  in  arithmetical  progression,  such 
that  the  product  of  the  extremes  shall  he  45,  and  the  product 
of  the  means  77. 

Let  a,  a  +  d,  a  +  2  d,  and  a  ±  3  d  he  the  quantities. 
Then,  by  the  conditions,  a2  ±  3  a  d  =  45  (1) 

a2  ±  3  a  d  +  2  d2  =  77  (2) 

Subtracting  (1)  from  (2),  2  d2  =  32 

d2  =  16 
t?=±4. 
If  d  =  4,  substituting  in  (1),  we  have 

a2  +  12a  =  45 

_                  -12±V^Ti4TT80      -12  ±18       Q 
Whence,  a  = —^ = ^ =  3  or  —  15. 

This  indicates  two  answers, 

3,  7,  11,  and  15,  or,  —  15,  —  11,  —  7,  and  —  3. 
If  d  =  —  4,  substituting  in  (1),  we  have 

a2— 12  a-  45 


W1  12±V144  +  180      12  ±18      1K  Q 

Whence,  a  = ! — ^ = „ =  15  or  —  3. 

This  also  indicates  two  answers, 

15,  11,  7,  and  3,  or,  —  3,  —  7,  —  11,  and  —  15. 

But  these  two  answers  are  the  same  as  those  obtained  with 
the  other  value  of  d.  Hence,  the  two  answers  to  the  problem 
are 

3,  7,  11,  and  15,  or,  —  3,  —  7,  — 11,  and  —  15. 


ARITHMETICAL  PROGRESSION.  281 

3.  Find  the  sum  of  the  odd  numbers  from  1  to  100. 

4.  A  debt  can  be  discharged  in  a  year  by  paying  $  1  the 
first  week,  $  3  the  second,  $  5  the  third,  and  so  on.  .Required 
the  last  payment,  and  the  amount  of  the  debt. 

5.  A  person  saves  $270  the  first  year,  $210  the  second, 
and  so  on.  In  how  many  years  will  a  person  who  saves  every 
year  $  180  have  saved  as  much  as  he  ? 

6.  Two  persons  start  together.  One  travels  ten  leagues  a 
day,  the  other  eight  leagues  the  first  day,  which  he  augments 
daily  by  half  a  league.  After  how  many  days,  and  at  what 
distance  from  the  point  of  departure,  will  they  come  together  ? 

7.  Find  four  numbers  in  arithmetical  progression,  such  that 
the  sum  of  the  first  and  third  shall  be  22,  and  the  sum  of  the 
second  and  fourth  36. 

8.  The  7th  term  of  an  arithmetical  progression  is  27  ;  and 
the  13th  term  is  51.     Find  the  first  term. 

9.  A  gentleman  set  out  from  Boston  to  NeW  York.  He 
travelled  25  miles  the  first  day,  20  miles  the  second  day,  each 
day  travelling  5  miles  less  than  on  the  preceding.  How  far 
was  he  from  Boston  at  the  end  of  the  eleventh  day  ? 

10.  If  a  man  travel  20  miles  the  first  day,  15  miles  the  sec- 
ond, and  so  continue  to  travel  5  miles  less  each  day,  how  far 
will  he  have  advanced  on  his  journey  at  the  end  of  the  8th 
day? 

11.  The  sum  of  the  squares  of  the  extremes  of  four  quanti- 
ties in  arithmetical  progression  is  200,  and  the  sum  of  the 
squares  of  the  means  is  13G.     What  are  the  quantities  ? 

12.  After  A  had  travelled  for  2|  hours,  at  the  rate  of  4 
miles  an  hour,  B  set  out  to  overtake  him,  and  went  4£  miles 
the  first  hour,  4J  the  second,  5  the  third,  and  so  on,  increasing 
his  speed  a  quarter  of  a  mile  every  hour.  In  how  many'  hours 
would  he  overtake  A  ? 


282  ALGEBRA. 


XXXIV.  —  GEOMETRICAL  PROGRESSION. 

374.  A  Geometrical  Progression  is  a  series  in  which  each 
term  is  derived  from  the  preceding  term  by  multiplying  by  a 
constant  quantity,  called  the  ratio. 

375.  When  the  series  is  increasing,  as,  for  exanrple. 

2,  6,  18,  54,  162,  

each  term  is  derived  from  the  preceding  term  by  multiplying 
by  a  quantity  greater  than  1  ;  consequently  the  ratio  is  a 
quantity  greater  than  1. 

When  the  series  is  decreasing,  as,  for  example, 
9   3   1    i    i    jl 

each  term  is  derived  from  the  preceding  term  by  multiplying 
by  a  quantity  less  than  1 ;  consequently  the  ratio  is  a  quantity 
less  than  1. 

Negative  values  of  the  ratio  are  admissible ;  for  example, 

-3,  6,-12,24,-48,  

is  a  progression  in  which  the  ratio  is  —  2. 

376.  Given  the  first  term,  a,  the  ratio,  r,  and  the  number 
of  terms,  n,  to  find  the  last  term,  I. 

The  progression  will  be 

a,  ar,  ar2,  ar5,  

We  observe  that  the  terms  differ  only  in  the  exponent  of  r, 
which  is  1  in  the  second  term,  2  in  the  third  term,  3  in  the 
fourth  term,  etc.  Consequently  in  the  nth.  or  last  term,  the 
exponent  of  r  will  be  n  —  1,  or 

l=arn-x  (1) 

377.  Given  the  first  term,  a,  the  last  term,  I,  and  the  ratio. 
r,  to  find  the  sum  of  the  series,  S. 


GEOMETRICAL  PROGRESSION.  283 

S=  a  +  a  r  +  a  r2  +  a  r3  + +  a  rn~3  +  a  r"-2  +  a  rn~x 

Multiplying  each  term  by  r, 

r  S—ar+  ar2+ar3+  a7A+ +  a  rn~2  +  a  r"_1  +  a  rn 

Subtracting  the  first  equation  from  the  second,  we  have 

ft    2,Tl   (f 

r  S—  S=arn  —  a,  or  S  (r  —l)  —  a  rn  —  a,  or  S  =  -     — ^— 
But  from  (1),  Art.  376,  by  multiplying  each  term  by  r, 

Substituting  this  value  of  a  rn  in  the  value  of  S,  we  have 

r  —  1 

378.     1.    In  the  series  2,  4,  8, to  11  terms,  find  the  last 

term  and  the  sum  of  the  series. 

Here  a  =  2,  n  =  ll;  the  ratio  is  always  found  by  dividing 

4 

the  second  term  by  the  first ;  hence,  r  =  -  =  2. 

Substituting  these  values  in  (1)  and  (2),  we  have 
I  =  2  (2)11-1  =  2  x  210  =  2  x  1024  =  2048. 
s=(2x204S)-2=4096_2  =  4094 

Zi  —  JL 

2.    In  the  series  3,  1.  »,  to  7  terms,  find  the  last  term 

o 

and  the  sum  of  the  series. 

Here  a  =  3,  n  =  7,  r  =  second  term  divided  by  first  term  =  ^. 

Substituting  these  values  in  (1)  and  (2),  we  have 
l-6\3)      ~*  \3)  _36_35_243" 


284  ALGEBEA. 

a   j_v  ,     r   ,      2186 

v3X243/      °    _729  "  729    _2186     3  _  1093 

^  =  ~^[  ~-   1  — "  ^2~     :  T29"  X  2  ~"  ^43" ' 

3_1  3  ~3 

3.    In  the  series  —  2,  6,  — 18,  to  8  terms,  find  the  last 

term  and  the  sum  of  the  series. 

Here  a  —  —  2,  n  =  S,  r=  — ~  =  —  3.     Hence, 

I  =  (_  2)  (-  3)8-1  =  (-  2)  (-  3)7  =  (-  2)  (-  2187)  =  4374. 

(_  3  x  4374)  -  (-  2)      -  13122  +  2  _^  -  13120  = 
(-3)  —  1-  -4  -4 

EXAMPLES. 

Find  the  last  term  and  the  sum  of  the  series  in  the  fol- 
lowing : 

4.  1,  2,  4, to  12  terms. 

4 

5.  3,  2,  - , to  7  terms. 

o 

6.  —2,  8,  —32, to  6  terms. 

7.  2,  —  1,  ■= , to  10  terms. 


111 

2'  V  8 


8.  ^ ,  T ,  q  , to  11  terms. 


2            3 
9.  k  i  —  1?  ^  i to  8  terms. 

10.  8,  4,  2, to  9  terms. 

11.  4' -4'  J2' to6  termS* 

2       11 

12.  -k>—  g>—  gj  to  10  terms. 

13.  3,  -6,  12, to  7  terms. 


GEOMETRICAL   PROGRESSION.  285 

379.  Formulae  (1)  and  (2)  together  contain  all  the  five  ele- 
ments of  a  geometrical  progression  ;  hence,  if  any  three  of  the 
five  are  given,  we  may  find  the  other  two,  exactly  as  in  arith- 
metical progression.  But  in  certain  cases  the  operation  in- 
volves the  solution  of  an  equation  of  a  higher  degree  than  the 
second,  for  which  rules  have  not  heen  given ;  and  in  other 
cases  the  unknown  quantity  appears  as  an  exponent,  the  solu- 
tion of  which  equations  can  usually  only  be  effected  by  the  use 
of  logarithms ;  although  in  certain  simple  cases  they  may  be 
solved  by  inspection. 

1.  Given  I  =  6561,  r  —  3,  n  =  9  ;  find  a  and  S. 

Substituting  these  values  in  (1)  and  (2),  Arts.  376  and  377, 
we  have 

6561  =  a  (3)8 ;  or  6561  =  6561  a  ;  or  a  =  1. 

(3x6561)-l_   19683-1  _  19682  _ 
6~         3=1  ~"   ~2~   "" ~2~ 

2.  Given  a  =  —  2,  n  =  5,  I  =  —  32  ;  find  r  and  S. 
Substituting  these  values  in  (1)  and  (2),  we  have 

-32  =  (-2)(r)5-1;  or-32  =  -2r1;  r4  =  16;  r  =  ±2. 

If  r  =  2,  S=  (2  X  ~32) 7  (~2)  =-  64  +  2  =  -62. 

T,  9    q     (-2x-32)-(-2)_64  +  2_66_ 

The  signification  of  the  two  answers  is  as  follows  : 

If  r=2,  the  series  will  be  -2,  -4,  -8,  -16,  -32,  in 
which  the  sum  is  —  62. 

If  r  =  —  2,  the  series  will  be  —  2,  4,  —  8,  16,  —  32,  in  which 
the  sum  is  —  22. 

3.  Given  a  =  3,  r  =  —  ^ ,  S  =  ;  find  n  and  I. 


286  ALGEBRA. 

Substituting  these  values  in  (1)  and  (2),  we  have 

-Us 

1640  _       3  1+9  •      6560  1 

"729—  ^TV=^r;  oW+9  =  T29  ;  °Tl  =  -729- 
3 

3 

Substituting  this  value  of  I  in  the  equation  (—  3)"_1  =  -,  we 

3 
have  (—  3)n~1  = ^-  =  —  2187;  whence,  by  inspection,  n— 1 


71".) 
=  7,  or  n  =  8. 

EXAMPLES. 

4.  Given  J  =  -  256,  r  =  —  2,  n  =  10 ;  find  a  and  A 

5.  Given  r  =  -,  n  =  8,  S—  -^-^r ;  find  a  and  Z. 

o  6561 

6.  Given  a  =  2,  n  =  7,  I  =  1458  ;  find  r  and  5. 

3 

7.  Given  a  =  3,  «  =  6,  Z  =  —  .  ,^ ,  ;  find  r  and  & 

1024 

8.  Given  a-=  1,  r  =  3,  Z  =  81 ;  find  -«  and  S. 

1  127 

9.  Given  a  =  2,  Z  =  ^  ,  $  =  -^- ;  find  n  and  r. 

10.  Given  «.  =  -  ,  r  =  —  3,  $=  —  91;  find  n  and  Z. 

11.  Given  £  =  -128,  r  =  2,  £  =  -255;  find  n  and  a. 

380.  The  Limit  to  which  the  sum  of  the  terms  of  a  decreas- 
ing geometrical  progression  approaches,  as  the  number  of  terms 
becomes  larger  and  larger,  is  called  the  sum  of  the  scries  to 
infinity.  We  may  write  the  value  of  S  obtained  in  Art.  .'177 
as  follows  : 


GEOMETRICAL  PROGRESSION.  287 

a  —  r  I 


S-. 


1-r 


In  a  decreasing  geometrical  progression,  the  larger  the  num- 
ber of  terms  taken  the  smaller  will  be  the  value  of  the  last 
term ;  hence,  by  taking  terms  enough,  the  last  term  may  be 
made  as  small  as  we  please.  Then  (Art.  207),  the  limiting 
value  of  I  is  0.  Consequently  the  limit  to  which  the  value  of 
S  approaches,  as  the  number  of  terms  becomes  larger  and 

larger,  is  - . 

Therefore  the  sum  of  a  decreasing  geometrical  progression 
to  infinity  is  given  by  the  formula 

1.  Find  the  sum  of  the  series  3,  1,  ^ ,  to  infinity. 

o 

Here  a  =  3,  r  =  k  ;  substituting  in  (3),  we  have 

O  3  9  9         A 

3 

8    16 

2.  Find  the  sum  of  the  series  4,  —  ^ ,  — ,   to  infinity. 

_8 

~~  3         2 
Here  a  =  4,  r  =  —r-  =  —  »  ;  substituting  in  (3),  we  have 

+  3 

EXAMPLES. 

Find  the  sum  of  the  following  to  infinity : 

3l   2'lj2' 5-   -1'3'~9' 


288 

ALGEBRA. 

3    11 

'■   4'  2'  3'     '■" 

9    8  2     1 
9-   *'5'  50' 

8    2   -A    li  in    -4  4    _A 

5'      35'  245' '  5'  "    25' 

381.     To  find  the  value  of  a  rejoeating  decimal. 

This  is  a  case  of  finding  the  sum  of  a  geometrical  progres- 
sion to  infinity,  and  may  be  solved  by  the  formula  of  Art.  380. 

1.   Find  the  value  of  .363636 

.363636 =  .36  +  .0036  +  .000036  + 


AAO/> 

Here  a  =  .36,  and  r  =  '—— — =  .01 ;  substituting  in  (3), 

.oo 

.36      _:36_36_  £ 

2.    Find  the  value  of  .285151  

.285151 =  .28  +  .0051  +  .000051  + 


To  find  the  sum  of  all  the  terms  except  the  first,  we  have 
a  =  .0051,  r  =  .01 ;  substituting  in  (3), 

a       .0051       .0051        51         17 

o  = 


1-.01        .99        9900     3300 

Adding  the  first  term  to  this,  the  value  of  the  given  decimal 

28         17         941 
~  100  +  3300"  3300' 


EXAMPLES. 

Find  the  values  of  the  following : 

3.  .074074 5.    .7333 7.    .113333. 

4.  .481481 6.   .52121 8.    .215454. 


GEOMETRICAL  PROGRESSION.  289 

382.     To  insert  any  number  of  geometrical  means  between 
two  given  terms. 

64 

1.  Insert  4  geometrical  means  between  2  and  p-r^ . 

This  may  be  performed  in  the  same  manner  as  the  examples 

64 
in  Art.  379.   We  have  a  =  2,  I  =  —-^ ,  and  n  =  6,  or  two  more 

243 

than  the  number  of  means. 

Substituting  these  values  in  (1),  Art.  376,  we  have 

64  32  2 

243  =  2?'5;  °rr5  =  243;  mr  =  t 

2 
Hence  the  terms  are  obtained  by  multiplying  2  by  ^  for  the 

2 
first,  that  result  by  ^  for  the  second,  and  so  on ;  or, 
o 

4    8    16    32    ^4 
2'3'  9'  27'  81'  243' 

2.  Insert  5  geometrical  means  between  —  2  and  — 128. 

Here  a  =  —  2,  I  =  — 128,  n  =  7.     Substituting  in  (1),  Art. 
376,  we  have 

—  128  =  —  2  7* ;   or  r6  =  64 ;  whence,  r  =  ±  2. 

If  r  =  2,  the  series  will  be 

_2;  -4,  -8,  -16,  -32,  -64,  -128. 

If  r  =  —  2,  the  series  will  be 

-  2,  4,  -  8,  16,  -  32,  64,  - 128. 


EXAMPLES. 

o  1    128 

3.  Insert  6  geometrical  means  between  o  and  ^Hq  • 

4.  Insert  5  geometrical  means  between  ^  and  364£. 

5.  Insert  6  geometrical  means  between  —  2  and  —  4374. 


200  ALGEBRA. 

729 
6.    Insert  4  geometrical  means  between  3  and  — 


3 


1024 


7.    Insert  7  geometrical  means  between  -  and 


PROBLEMS. 

383.     1.    What  is  the  first  term  of  a  geometrical  progression, 
when  the  5th  term  is  48,  and  the  8th  term  is  —  384  ? 

By  Art.  376,  the  5th  term  is  a  r4,  and  the  8th  term  is  a  r~. 

Hence, 

ar*  =  48,  and  a  r'  =  —  384. 

Dividing  the  second  of  these  equations  by  the  first, 
r3  =  —  8  ;  whence,  r  =  —  2. 

Tl  48       48        <*     J 

Ihen,  «  =  — ; -  =  :r7r=z:  3 ,  Ans. 

r*       16 

2.    Find  three  numbers  in  geometrical  progression,  such  that 
their  sum  shall  be  14,  and  the  sum  of  their  squares  84. 

Let  "a,  a  r,  and  a  r~  denote  the  numbers.     Then,  by  the  con- 
ditions, 

a  +  a  r  +  a  r2  =  14  (1) 

a2  +  a2r2  +  a2ri  =  8A  (2) 

Dividing  (2)  by  (1),  a  —  ar+ar2  =  6  (3) 

Adding  (1)  and  (3),  a  +  a  r2  =  10  (4) 

4 

Subtracting  (3)  from  (1),  a  r  =  4,  ofr  =  -        (5) 

1  c 

Substituting  from  (5)  in  (4),  a  -\ =  10 


<r 


-10  a  =  -16 


wi           (K       onox           lOiy/100-64      10  ±6      Q       _ 
VV  hence   (Art.  309),  a  — ~ -  =  — - —  —  8  or  2. 

—  Z 

4      1 
If  a  =  8,  r  =  q  =  jr,  and  the  numbers  are  8,  4,  and  2. 

8      Z 


HARMONICAL   PROGRESSION.  291 

4 
If  a  =  2,  ?•  =  -  =  2,  and  the  numbers  are  2,  4,  and  8. 

Therefore,  the  numbers  are  2,  4,  and  8,  Ans. 

3.  A  person  who  saved  every  year  half  as  much  again  as  he 
saved  the  previous  year,  had  in  seven  years  saved  $2059. 
How  much  did  he  save  the  first  year  ? 

4.  A  gentleman  boarded  9  days,  paying  3  cents  for  the  first 
day,  9  cents  for  the  second  day,  27  cents  for  the  third  day,  and 
so  on.     Required  the  cost. 

5.  Suppose  the  elastic  power  of  a  ball  that  falls  from  a 
height  of  100  feet,  to  be  such  as  to  cause  it  to  rise  0.9375  of 
the  height  from  which  it  fell,  and  to  continue  in  this  way 
diminishing  the  height  to  which  it  will  rise,  in  geometrical 
progression,  till  it  comes  to  rest.     How  far  will  it  have  moved  ? 

6.  The  sum  of  the  first  and  second  of  four  quantities  in 
geometrical  progression  is  15,  and  the  sum  of  the  third  and 
fourth  is  60.     Required  the  quantities. 

7.  The  fifth  term  of  a  geometrical  progression  is  —  324,  and 
the  9th  term  is  —  26244.     What  is  the  first  term  ? 

8.  The  third  term  of  a  geometrical  progression  is  ^-r ,  and 

9 
the  sixth  term  is  -^r^ .     What  is  the  second  term  ? 


XXXV.— HARMONICAL  PROGRESSION. 

384.     Quantities  are  said  to  be  in  Harmonical  Progression 
when  their  reciprocals  form  an  arithmetical  progression. 

For  example,  1,  -,  F,  -,  


3'  5'  7 
are  in  harmonical  progression,  because  their  reciprocals, 

1,  3,  5,  7,  

form  an  arithmetical  progression. 


292  ALGEBRA. 

385.  From  the  preceding  it  follows  that  all  problems  in 
harmonical  progression,  which  are  susceptible  of  solution,  may 
be  solved  by  inverting  the  terms  and  applying  the  rules  of  the 
arithmetical  progression.  There  will  be  found,  however,  no 
general  expression  for  the  sum  of  a  harmonical  series. 

386.  To  find  the  last  term  of  a  given  ha  rmonical  series. 


2    2 

3'  5; 


1.    In  the  series  2,  - ,  p, to  36  terms,  find  the  last  term. 


Inverting  the  series,  we  have  the  arithmetical  progression 

2  j  2 '  2 '  t0  36  terms• 

Here  a  =  -,  d  =  l,  w  =  36 ;  hence,  by  (1),  Art.  368, 

Z  =  i+(36-l)l  =  ^  +  35  =  ^. 

2 
Inverting  this,  we  obtain  -=r  as  the  last  term  of  the  given  series. 


EXAMPLES. 

Find  the  last  terms  of  the  following : 

K       Q  A      *}      1 9 

2.  - ,  - ,  to  23  terms.  4.  r,T,-, to  26  terms. 

3.  -  ---r, to  17  terms.    5.  a,  b,  to  n  terms. 

2      3      o 

387.     To  ii/si'rf  any  number  of  harmonical  means  between 
two  given  terms. 

1.    Insert  5  harmonical  means  between  2  and  — 3. 
Inverting,  we  have  to  insert  5  arithmetical  means  between 

1        A        1 

2and~3' 


HARMONICAL  PROGRESSION.  293 

Here  a  =  - ,  Z  =  —  -,n  =  J;  substituting  in  (1),  Art,  368, 
Z  o 

we  have 

—  -  —  -  +  6d:  or  6  d  —  —  -x :  whence,  cZ  =  —  —  . 
3      2  6  ob 

Hence,  the  arithmetical  means  are 

13    2     1_        _1_        _L 
36' 9' 12'      18'      36* 

Then,  the  harmonical  means  will  be 

36    9    ._       1Q        36 

-,-,12,-lS, -T,Ans. 

EXAMPLES. 

2  3 

2.  Insert  7  harmonical  means  between  -  and  — . 

5  10 

3.  Insert  3  harmonical  means  between  —  1  and  —  5. 

4.  Insert  6  harmonical  means  between  3  and  —  1. 

5.  Insert  m  harmonical  means  between  a  and  b. 

388.  If  three  consecutive  terms  of  a  harmonical  progres- 
sion be  taken,  the  first  has  the  same  ratio  to  the  third,  that 
the  first  minus  the  second  has  to  the  second  minus  the  third. 

Let  a,  b,  c  be  in  harmonical  progression ;  then  their  recip- 
rocals -  ,  - ,  and  -  will  be  in  arithmetical  progression.    Hence^ 
a>    o  c 

1_1_1_1 

c       b      b       a' 

Clearing  of  fractions,  ab  —  ac  =  ac  —  bc 
or,  a  (b  —  c)  =  c  (a  —  b) 

Dividing  through  by  c  (b  —  c),  we  have 

a      a  —  b 

which  was  to  be  proved. 


294  »  ALGEBRA. 

389.  Let  a  and  c  be  any  two  quantities  ;  b  their  harrnoni- 
cal  mean.     Then,  by  the  previous  theorem,  -  = . 

Clearing  of  fractions,  ab  —  ac  =  ac  —  be;  then,  ab  +  bc  =  2ac 

2  a  c 
or,  b  = . 

a  +  c 

390.  We  may  note  the  following  results :  if  a  and  c  arc 

a  ~f~  c 
any  two  quantities,  their    arithmetical  mean  =  — - — ;  their 

geometrical  mean  =  \a  c ;  and  their  harmonical  mean  =  -     -  . 

a+  c 

a-         2  a  c       a  +  c        /  / — \2    . 

Since X  — 7. —  =  vV  a  c)  >  lt  follows  that  the  product 

a  -\-  c  ~j 

of  the  harmonical  and  arithmetical  means  of  two  quantities  is 

equal  to  the  square  of  their  geometrical  mean. 

Consequently  the  geometrical  mean  must  be  intermediate  in 

value  between  the  harmonical  and  the  arithmetical  mean.     But 

the  harmonical  mean  is  less  than  the  arithmetical  mean,  be- 

a  +  c        2ac       (a  +  e)2  —  4  a  c      a2 +  2  ac+  c2  —  £ac 

cause  — pr =  - — T—/- r = — — r 

2  a  +  c  2  (a  +  c)  2  (a  +  c) 

a2  —  2ac  +  c2       (a  —  c)2 

=  — 7T7 ^ =  7T^ n  a  positive  quantity. 

2  (a  +  c)  2  (a  +  c)  '     L  >■  * 

Hence  of  the  three  quantities,  the  arithmetical  mean  is  the 
greatest,  the  geometrical  mean  next,  and  the  harmonical  mean 
the  least. 


XXXVI.  — PERMUTATIONS  AND  COMBINA- 
TIONS. 

391.     The  different  orders  in  which  quantities  can  be  ar- 
ranged are  called  their  Permutations. 

Thus,  the  permutations  of  the  quantities  a,  b,  <;  taken  two 
at  a  time,  are 


PERMUTATIONS   AND   COMBINATIONS.  295 

a  b,  b  a  ;  a  c,  c  a;  be,  c  b; 
and  taken  three  at  a  time,  are 

a  b  c,  ac  b;  b  a  e,  b  c  a  ;  cab,  cb  a. 

392.  The  Combinations  of  quantities  are  the  different  col- 
lections that  can  he  formed  out  of  them,  without  regard  to  the 
order  in  which  they  are  placed. 

Thus,  the  combinations  of  the  quantities  a,  b,  c,  taken  two 

at  a  time,  are 

a  b,  ac,  be; 

a  b,  and  b  a,  though  different  permutations,  forming  the  same 
combination. 

393.  To  find  the  number  of  permutations  of  n  quantities, 

taken  r  at  a  time. 

Let  P  denote  the  number  of  permutations  of  n  quantities, 
taken  r  at  a  time.  By  placing  before  each  of  these  the  other 
n  —  r  quantities  one  at' a  time,  we  shall  evidently  form  P  (n  —  r) 
permutations  of  the  n  quantities,  taken  r  +  1  at  a  time.  That 
is,  the  number  of  permutations  of  n  quantities,  taken  r  at  a 
time,  multiplied  by  n  —  r,  gives  the  number  of  j)ermutations 
of  the  n  quantities,  taken  r  +  1  at  a  time. 

But  the  number  of  permutations  of  n  quantities,  taken  one 
at  a  time,  is  obviously  n.  Hence,  the  number  of  permutations 
taken  two  at  a  time,  is  the  number  taken  one  at  a  time, 
multiplied  by  n  —  1,  or  n  (n—  1).  The  number  of  permuta- 
tions, taken  three  at  a  time,  is  the  number  taken  two  at  a  time, 
multiplied  by  n  —  2,  or  n  (n  —  1)  (n  —  2);  and  so  on.  We 
observe  that  the  last  factor  in  the  number  of  permutations  is 
n,  minus  a  number  1  less  than  the  number  of  quantities  taken 
at  a  tinie.  Hence,  the  number  of  permutations  of  n  quanti- 
ties, taken  r  at  a  time,  is 

n(n-l)  0-2)  (n-(r-l)) 

or,  n(n  —  l)(n  —  2) (n  —  r  +  T).  (1) 


296  ALGEBRA. 

394.  If  all  the  quantities  are  taken  together,  r  =  n  and 
Formula  (1)  becomes 

n(n-l)  (n-2) 1; 

or,  by  inverting  the  order  of  the  factors, 

1x2x3 (n— l)n.  (2) 

That  is, 

The  number  of  permutations  of  n  quantities,  taken  n  at  a 
time,  is  equal  to  the  product  of  the  natural  numbers  from  1 
up  to  n. 

For  the  sake  of  brevity,  this  result  is  often  denoted  by  \n, 
read  "factorial  n  "  ;  thus  \n_  denotes  the  product  of  the  natu- 
ral numbers  from  1  to  n  inclusive. 

395.  To  find  the  number  of  combinations  of  n  quantities, 
taken  r  at  a  time. 

The  number  of  permutations  of  n  quantities,  taken  r  at  a 
time,  is  (Art.  393), 

n{n  —  l)  (n  —  2)  (n  —  r+  1). 

By  Art.  394,  each  combination  of  r  quantities  produces  \r 

permutations.     Hence,  the  number  of  combinations  must  equal 
the  number  of  permutations  divided  by  ta  or 

n(n-l)  (n-2) (n-r+1) 

|  r 

396.  The  number  of  combinations  of  n  quant  It  Its,  taken  r 
at  a  time  is  the  same  as  the  number  of  combinations  of  n 
quantities  taken  n  —  r  at  a  time. 

For,  it  is  evident  that  for  every  combination  of  r  quantities 
which  we  take  out  of  n  quantities,  we  leave  one  combination 
of  n  —  r  quantities,  which  contains  the  remaining  quantities. 

EXAMPLES. 

397.  1.  How  many  changes  can  be  rung  with  10  bells, 
taking  7  at  a  time  ? 


PERMUTATIONS   AND   COMBINATIONS.  2! >7 

Here,  n  =  10,  r  =  7  ;  then  n  —  r  +  1  =  4. 

Then,  by  Formula  (1), 

10x9x8x7x6x5x4  =  604800,  ^».s. 

2.    How  many  different  combinations  can  be  made  with  5 
letters  out  of  8  ? 

Here,  n  =  8,  r  =  5  ;  then  n  —  r  +  1  =  4:. 

Then,  by  Formula  (3), 

8x7x6x5x4 


1x2x3x4x5 


56,  Ans. 


3.    In  how  many  different  orders  may  7  persons  be  seated 
at  table  ? 

Here  n  =  7 ;  then,  by  Formula  (2), 

1x2x3x4x5x6x7  =  5040,  Ans. 

4.  How  many  different  words  of  4  letters  each  can  be  made 
with  6  letters  ?  How  many  words  of  3  letters  each  ?  How 
many  of  6  letters  each  ?     How  many  in  all  possible  ways  ? 

5.  How  often  can  4  students  change  their  places  in  a  class 
of  8,  so  as  not  to  preserve  the  same  order  ? 

6.  From  a  company  of  40  soldiers,  how  many  different  pick- 
ets of  6  men  can  be  taken  ? 

7.  How  many  permutations  can  be  formed  of  the  26  letters 
of  the  alphabet,  taken  4  at  a  time  ? 

8.  How  many  different  numbers  can  be  formed  with  the 
digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  taking  5  at  a  time,  each  digit 
occurring  not  more  than  once  in  any  number  ? 

9.  How  many  different  permutations  may  be  formed  of  the 
letters  in  the  word  since,  taken  all  together  ? 

10.    How  many  different  combinations  may  be  formed  of  the 
letters  in  the  word  forming,  taken  three  at  a  time  ? 


298  ALGEBRA. 

11.  How  many  different  combinations  may  be  formed  of  20 
letters,  taken  5  at  a  time  ? 

12.  How  many  different  combinations  may  be  formed  of  18 
letters,  taken  11  at  a  time  ? 

13.  How  many  different  committees,  consisting  of  7  persons 
each,  can  be  formed  out  of  a  corporation  of  20  persons  ? 

14.  How  many  different  numbers,  of  three  different  figures 
each,  can  be  formed  from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  0  ? 


XXXVII.  —  BINOMIAL  THEOREM. 

POSITIVE  INTEGRAL  EXPONENT. 

398.  The  Binomial  Theorem,  discovered  by  Newton,  is  a 
formula,  by  means  of  which  any  binomial  may  be  raised  to  any 
required  power,  without  going  through  the  process  of  invo- 
lution. 

399.  To  prove  the  Theorem  for  a  positive  integral  ex- 
ponent. 

By  actual  multiplication  we  may  show  that 
(a  +  x)2  =  a2  +  2  ax  +  x2 
(a  +  x)3  =  a3  +  3  a2  x  +  3  a  x2  +  x3 
(a  +  x)4  =  a4  +  4  a3  x  +  6  a2  x2  +  4  a  x3  +  x* 

In  these  results  we  observe  the  following  laws : 

1.  The  number  of  terms  is  one  more  than  the  exponent  of  the 
binomial. 

2.  The  exponent  of  a  in  the  first  term  is  the  same  as  the 
exponent  of  the  binomial,  and  decreases  by  one  in  each  stic- 
ceedin;/  term. 

3.  The  exponent  of  x  in  the  second  term  is  unity,  and  in- 
creases by  one  in  each  succeeding  term. 


BINOMIAL  THEOREM.  299 

4.  The  coefficient  of  the  first  term,  is  unity ;  and  of  the  sec- 
ond term,  is  the  exponent  of  the  binomial. 

5.  If  the  coefficient  of  any  term  be  multiplied  by  the  expo- 
nent of  a  in  that  term,  and  the  product  divided  by  the  number 
of  the  term,  beginning  at  the  left,  the  result  will  be  the  co- 
efficient of  the  next  term,. 

Assuming  that  the  laws  hold  for  any  positive  integral  expo- 
nent, nr  we  have 

.         .                      ,       n(n— 1)       „  „    n(n— T)(n— 2)       „  , 
(a+x)n=an+nan-1x+   \       'an~2x2+    K       J±- /an~sx3+ 

This  result  is  called  the  Binomial  Theorem. 

400.  To  prove  that  it  holds  for  any  positive  integral  expo- 
nent, we  multiply  hoth  members  by  a  +  x,  thus 

/         x  ^i          , ,                  n(n—T)       ,    ,     n(n — l)(n— 2)       .  „ 
(a+x)n+1=an+1+?ianx  +    \      V^ar^-    v         J\ ' an~2x3 

,    „     n(n— 1)        .    , 


+ +  anx  +  nan  ~yx~+    v  an~ix6+. 

X.A 


=  an  +  1+  (n  +  1)  an  x  + 


n  (n  —  1) 


a""1  x- 


ln{n-l)(n-2)      ^-1)1 
+  L        1.2.3        "+      1.2     J  +> 

=  an+1  +  (w  + 1)  an  x  +  -™2  In  - 1  +  2 1  an~l  a? 

+nS  [-H  <-*+ 

{  ?i  "T~  1  )    72- 

=  a"+1  +  (n  + 1)  a"  a;  +  ^-j-tj      a"-1  ar2 


(w  +  1)  n  (n—  1)     _„    3 
1.2.3       ~a"  "  *  + 


where  it  is  evident  that  every  term  except  the  first  will  con- 
tain the  factor  n  + 1. 


300  ALGEBRA. 

"We  observe  that  the  expansion  is  of  the  same  form  as  the 
value  of  (a  +  x)n,  having  n  +  1  in  the  place  of  n. 

Hence,  if  the  laws  of  Art.  399  hold  for  any  positive  integral 
exponent,  n,  they  also  hold  when  that  exponent  is  increased 
by  1.  But  we  have  shown  them  to  bold  for  (a  +  cc)4,  hence 
they  hold  for  (a  +  x)5;  and  since  they  hold  for  (a  +  x)5,  they 
also  hold  for  (a  +  x)6;  and  so  on.  Hence  they  hold  for  any 
positive  integral  exponent. 

401.  Since  1.2  =  [2,  1.  2. 3  =  [3,  etc.  (Art.  394),  the  Bino- 
mial Theorem  is  usually  written  as  follows : 

\      n(n—l)      „  o    n(n—l)(n—2)  _  ,  „ 
(a+x)n=an+?i  an  ~lx+   V       V~V+— £ }  an^x3+ 

If.  £L 

402.  If  a  =1,  then,  since  any  power  of  1  equals  1,  we 
bave 

?i(n  —  l)    ,     w  (n.  —  1)  (w  —  2)    „ 

403.  In  performing  examples  by  the  aid  of  the  Binomial 
Theorem,  we  may  use  the  laws  of  Art.  399  to  find  the  expo- 
nents and  coefficients  of  the  terms. 

1.   Expand  (a  +  x)6  by  the  Binomial  Theorem. 

The  number  of  terms  is  7. 

The  exponent  of  a  in  the  first  term  is  6,  and  decreases  by  1 
in  each  succeeding  term. 

The  exponent  of  x  in  the  second  term  is  1,  and  increases  by 
1  in  each  succeeding  term. 

The  coefficient  of  the  first  term  is  1 ;  of  the  second  term,  6; 
if  the  coefficient  of  the  second  term,  6,  be  multiplied  by  5,  the 
exponent  of  a  in  that  term,  and  the  product,  30,  be  divided  by 
the  number  of  the  term,  2,  the  result,  15,  will  be  the  coefficient 
of  the  third  term  ;  etc. 

Eesult,  a6+6a5x  +  15  a4  x2  +  20  a3  x*  +  15  a2  x"  +  6  a  x5  +  x*. 


BINOMIAL   THEOREM.  301 

Note.  It  will  be  observed  that  the  coefficients  of  any  two  terms  taken 
equidistant  from  the  beginning  and  end  of  the  expansion  are  the  same. 
The  reason  for  this  will  be  obvious  if,  in  Art.  401,  x  and  a  be  inter- 
changed, which  is  equivalent  to  inverting  the  series  in  the  second  member. 
Thus,  the  coefficients  of  the  latter  half  of  an  expansion  may  bo  written 
out  from  the  first  half. 

2.  Expand  (1  +  sc)7  by  the  Binomial  Theorem.     " 

Result,  l7  +  7.16.  x  +  21.15.  x2  +  35.14.  x3  +  35.13.  x*  +  21.1s.  x5 
+  7.11.  x°  +  x1  ; 

or,     1  +  7  x  +  21  x2  +  35  x3  +  35  xA  +  21  x5  +  7  x6  +  x\ 

Note.  If  the  first  term  of  the  binomial  is  a  numerical  quantity,  it  will 
be  found  convenient,  in  applying  the  laws,  to  retain  the  exponents  at  first 
without  reduction,  as  then  the  laws  for  coefficients  may  be  used.  The  re- 
sult should  afterwards  be  reduced  to  its  simplest  form. 

3.  Expand  (2  a  +  3  b)'°  by  the  Binomial  Theorem. 

(2a  +  3&)5=[(2«.)  +  (36)]5 

=  (2  a)5  +  5  (2  a)4  (3  b)  +  10  (2  a)3  (3  bf  +  10  (2  a)'2  (3  b)3 
+  5  (2  a)  (3  by  +  (3  b)5 

=  32  a5  +  240  a4  b  +  720  a3  V2  +  1080  a2  b3  +  810  a  b*  +  243  b5, 

Aiis. 

4.  Expand  (irf2  —  ft-1)6  by  the  Binomial  Theorem. 

(m-i  _  n-i)«  =  [(m_i)  +  (-  ft-1)]6 

=(m-^)0+6(m"2)5(_w-i)+i5(m-*)4(-«-1)'2+20(m"^)3(-H-1)3 
+ 15  (m~ty  (-ft"1)4  +  G  (m~*)  (-n-lf+  (-ft-1)6  ' 

=mr3  +  6  m~r%  (-  ft"1)  +  15  m~2  (ft-2)  +  20  i»~*  (-  ft"3) 
+  15  m-1  (ft-4)  +  6  m~-  (-  ft"5)  +  (ft"6) 

=  m~3  —  6  m~*  ft-1  +  15  m~-  n~-  —  20  m~  -  ft~3  +  15  m~x  ft"4 
—  6  m   "2  ft-5  +  ft-6,  Ans. 


302  ALGEBRA. 

Note.  If  either  term  of  the  binomial  is  not  a  single  letter,  with  unity 
as  its  coefficient  and  exponent,  or  if  either  term  is  preceded  by  a  minus 
sign,  it  will  be  found  convenient  to  enclose  the  term,  sign  and  all,  in  a 
parenthesis,  when  the  usual  laws  for  exponents  and  coefficients  may  be 
applied.  In  reducing,  care  must  be  taken  to  apply  the  principles  of  Arts. 
227  and  259. 

EXAMPLES. 

Expand  the  following  by  the  Binomial  Theorem : 


5. 

(1  +  o)5. 

8. 

(a  b  —  c  d)~. 

11. 

(J  +  <$y. 

6. 

(a  +  a-3)6. 

9. 

O2  +  3  nr)\ 

12. 

(m~%  +  2  ?isy. 

7. 

(x*-2yy. 

10. 

(a-2  -  4  x*)b. 

13. 

(a-i  — &»aj4)«. 

404.     To  find  the  rth  or  general  term  of  the  expansion  of 

(a  +  ,r)\ 

"We  have  now  to  determine,  from  the  observed  laws  of  the 
expansion,  three  things ;  the  exponent  of  a  in  the  term,  the 
exponent  of  x  in  the  term,  and  the  coefficient  of  the  term. 

The  exponent  of  x  in  the  second  term  is  1 ;  in  the  third 
term,  2  ;  etc.     Hence,  in  the  rth  term  it  will  be  r  —  1. 

In  any  term  the  sum  of  the  exponents  of  a  and  x  is  n. 
Hence,  in  the  rth  term,  the  exponent  of  a  will  be  such  a  quan- 
tity as  when  added  to  r  —  1,  the  exponent  of  x,  will  produce 
n  ;  or,  the  exponent  of  a  will  be  n  —  r  +  1. 

The  coefficient  of  the  term  will  be  a  fraction,  of  the  form 

n(n-l)(»-2) ,  .  ,  ,  .       ,     , 

-. — -} — q—    ;  m  which  we  must  determine  the  last 

X-  .  w  .  o 

factors  of  the  numerator  and  denominator. 

We  observe  that  the  last  factor  of  the  numerator  of  any 
tennis  1  more  than  the  exponent  of  a  in  that  term:  hence 
the  last  factor  of  the  numerator  of  the  rth  term  will  be 
n  —  r  +  2. 

Also,  the  last  factor  of  the  denominator  of  any  term  is  the 
same  as  the  exponent  of  x  in  that  term  :  hence  the  last  factor 
of  the  denominator  of  the  rth  term  will  be  r  —  1. 


BINOMIAL   THEOREM.  303 

Therefore  the 

n  (n  -  1)  (n  -  2) (n  —  r  +  2) 

rth  term  =  -^ t^t;-^ — ^^^rr  -1—  ctn~r  +  1  xr~\ 

1 .  Z  .  6 (r  —  1) 

1.    Find  the  8th  term  of  (3  J  -2b~l)n. 
Here  r  =  8,  n  =  11 ;  hence,  the 

8thtom=1l:2°39485.6675<3a*>,(-2t-)' 

=  330  (81  a2)  (-  128  b~'')  =  -  3421440  a2  J-7,  ^ras. 

Note.     The  note  to  Ex.  4,  Art.  403,  applies  with  equal  force  to  examples 
in  this  article. 


EXAMPLES. 

Find  the 

2.  10th  term  of  (a  +  x)15.  5.    5th  term  of  (1  -  a2)12. 

3.  6th  term  of  (1  +  m)u.  6.    9th  term  of  (x~1-2  y*)u. 

4.  8th  term  of  (c  -  d)l\  7.    8th  term  of  (a%  +  3  x"1)10. 

405.  A  trinomial  may  he  raised  to  any  power  by  the  Bi- 
nomial Theorem,  if  two  of  its  terms  he  enclosed  in  a  paren- 
thesis and  regarded  as  a  single  term  ;  the  operations  indicated 
being  performed  after  the  expansion  by  the  Theorem  has  been 
effected. 

1.    Expand  (2  a  —  b  +  c2)3  by  the  Binomial  Theorem. 

(2  a -  b  +  c2)3  =  [(2  a-b)  +  (c2)]3 

=  (2  a  -  by  +  3  (2  a -b)2  (c2)  +  3  (2  a  -  b)  (c2)2  +  (c2)3 

=Sa3~12a2b  +  6ab2-b3+3c2(4:a2-4ab  +  b2)  +  3c4(2a-b)  +  c6 

=  8  a3  -  12  a2  b  +  6  a  b2  -  b3  +  12  a2  c2  -  12  a  b  c2  +  3  b2  c2 
+  6aci-3bc4  +  c6,  Ans. 


304  ALGEBRA. 

The  same  method  will  apply  to  the  expansion  of  any  poly- 
nomial hy  the  Binomial  Theorem. 

EXAMPLES. 

Expand  the  following  by  the  Binomial  Theorem  : 

2.  (1-x-x2)*.  4.    (1  -  2  x  -  2  x2)\ 

3.  (x-+3x  +  l)3.  5.    (l  +  a;-a;2)5. 


XXXVIII.— UNDETERMINED  COEFFICIENTS. 

406.  A  Series  is  a  succession  of  terms,  so  related  that  each 
may  be  derived  from  one  or  more  of  the  others,  in  accordance 
with  some  fixed  law. 

The  simpler  forms  of  series  have  already  been  exhibited  in 
the  progressions. 

407.  A  Finite  Series  is  one  having  a  finite  number  of 
terms. 

408.  An  Infinite  Series  is  one  whose  number  of  terms  is 
unlimited. 

The  progressions,  in  general,  are  examples  of  finite  series ; 
but,  in  Art.  380,  we  considered  infinite  Geometrical  series. 

409.  An  infinite  series  is  said  to  be  convergent  when  the 
sum  of  the  first  n  terms  cannot  numerically  exceed  some  finite 
quantity,  however  large  n  may  be ;  and  it  is  said  to  be  diver- 
gent when  the  sum  of  the  first  n  terms  can  numerically  exceed 
any  finite  quantity  by  taking  n  large  enough. 

For  example,  consider  the  infinite  series 

1  +  x  +  x-  +  X3  + 

The  sum  of  the  first  n  terms 

1  +  x  +  x2  +  x3+ +xn~1  =  ^^     (Art.  120). 

1  —  x 


UNDETERMINED  COEFFICIENTS.  305 

If  x  is  less  than  1,  xn  is  less  than  x,  however  largo  n  may 
be ;  consequently  the  numerator  and  denominator  of  the  frac- 
tion are  each  less  than  1,  and  positive ;  and  the  numerator  is 
larger  than  the  denominator ;  hence  the  fraction  is  equal  to 
some  finite  quantity  greater  than  1.  The  series  is  therefore 
convergent  if  x  is  less  than  1. 

If  x  is  equal  to  1,  each  term  of  the  series  equals  1,  conse- 
quently the  sum  of  the  first  n  terms  is  n ;  and  this  can  numer- 
ically exceed  any  finite  quantity  by  taking  n  large  enough. 
The  series  is  therefore  divergent  if  x  =  1. 

If  x  is  greater  than  1,  each  term  of  the  series  after  the  first 
is  greater  than  1,  consequently  the  sum  of  the  first  n  terms  is 
greater  than  n  ;  and  this  sum  can  numerically  exceed  any  finite 
quantity  by  taking  n  large  enough.  The  series  is  therefore 
divergent  if  x  is  greater  than  1. 

410.  Every  infinite  literal  series,  arranged  in  order  of  pow- 
ers of  some  letter,  is  convergent  for  some  values  of  that  letter, 
and  divergent  for  other  values. 

We  will  now  show  that  it  is  convergent  when  that  letter 
equals  zero. 

Let  the  series  be 

a  +  bx  +  cx2  +  dxs+ +  Jexn~1+ 

The  sum  of  the  first  n  terms  is 

a  +  bx  +  cx2  +  dx3+ +  kxn~\ 

which  is  equal  to  a,  if  x  is  made  equal  to  0. 

Hence,  however  large  n  may  be,  the  sum  of  the  first  n  terms 
is  equal  to  a,  if  x  is  equal  to  0.  Consequently  the  series  is 
convergent  if  x  =  0. 

411.  Infinite  series  may  be  developed  by  the  common  pro- 
cesses of  Division,  as  in  Art.  101,  Exs.  19  and  20,  and  Extrac- 
tion of  Roots,  as  in  Arts.  239  and  243 ;  and  by  other  methods 
which  it  will  now  be  our  object  to  elucidate. 


306  ALGEBRA. 


UNDETERMINED   COEFFICIENTS. 

412.  A  method  of  expanding  algebraic  expressions  into 
series,  simple  in  its  principles,  and  general  in  its  application, 
is  based  on  the  following  theorem,  known  as  the 


THEOREM  OF  "UNDETERMINED  COEFFICIENTS. 

413.     If  the  series  A  +  Bx  +  Cx2  +  Bx3  + is  always 

equal  to  the  series  A'  +  B'  x  -f-  C  ar  -f-  D'  x3  + ,  for  a  ny 

value  of  x  which  makes  both  series  convergent,  the  coefficients 
of  like  powers  of ' x  in  the  two  series  will  he  equal. 

For,  since  the  equation 
A  +  Bx  +  Cx2+Dx3  + =A'+B'x+  C'x-+D>x3  + 


is  satisfied  for  any  value  of  x  which  makes  both  members  con- 
vergent ;  and  since  by  Art.  410,  if  x  is  equal  to  0,  both  mem- 
bers are  convergent ;  it  follows  that  the  equation  is  satisfied  if 
x  =  0.     Making  x  —  0,  the  equation  becomes 

A  =  A'. 

Subtracting  A  from  the  first  member  of  the  equation,  and  its 
equal,  A',  from  the  second  member,  we  have 

Bx+  Cx2+Bx3  + =  B'x  +  C'x2  +  D'x3  + 

Dividing  through  by  x, 

B+Cx  +  Bx2+ =B'+  C'x  +  D'x2+ 


This  equation  is  also  satisfied  for  any  value  of  X  which  makes 
both  members  convergent ;  hence  it  is  satisfied  if  x  =  0.  Mak- 
ing x  =  0,  we  have; 

B  =  B'. 

Proceeding  in  this  way,  we  may  show  C=  O,  I)  =  D',  etc. 


UNDETERMINED   COEFFICIENTS.  307 

Note.  The  necessity  for  the  limitation  of  the  theorem  to  values  of  x 
which  make  both  series  convergent,  is  that  a  convergent  series  evidently 
cannot  be  equal  to  a  divergent  series  ;  and  two  divergent  scries  cannot  be 
equal,  as  two  quantities  which  numerically  exceed  any  finite  quantity  can- 
not be  said  to  be  equal. 

Hence,  in  all  applications  of  the  theorem,  the  results  are  only  true  when 
both  members  are  convergent. 


APPLICATION  TO  THE  EXPANSION  OF  FRACTIONS  INTO  SERIES. 

2  +  5  x 

414.     1.    Expand  - — -  into  a  series. 

JL  —  o  x 

We  have  seen  (Art.  101),  that  any  fraction  may  he  expanded 
into  a  series  by  dividing  the  numerator  by  the  denominator ; 
consequently,  we  know  that  the  proposed  development  is  pos- 
sible.    Assume  then, 

^  +  f  x  =A+Bx  +  Cx2  +  Dx*+Exi+ (1) 

1  —  ox 

where  A,  B,  C,  D,  E,  are  quantities  independent  of  x. 

Clearing  of  fractions,  and  collecting  together  in  the  second 
member  the  terms  containing  like  powers  of  x,  we  have 


2+5x  =  A  +    B 
-3.4 


x+     C 
-3B 


x-+    D 
-3(7 


x3  +     E 
-3D 


x*  + 


Equation  (1),  and  also  the  preceding  equation,  are  evidently 
to  be  satisfied  by  all  values  of  x  which  make  the  second  mem- 
ber a  convergent  series.  Hence,  applying  the  Theorem  of  Un- 
determined Coefficients  to  the  latter,  we  have 

A  =  2. 
B  —  3  A  —  5 ;  whence,  B  =  3  A  +  5  =  11. 
C-3B  =  0;  whence,  C  =  3B         =33. 
D  -  3  C  =  0  ;  whence,  D  =  3  C        =  99. 
E-3D  =  0;  whence,  E=3D        =297. 


303 


ALGEBRA. 


Substituting  these  values  of  A,  B,  C,  D,  E,  in  (1),  we 

have 

^+-^  =  2  +  11  a:  +  33  z2  +  99  cc3  +  297  a:4  +  , 

1  —  o  x 

which  may  he  readily  verified  by  division. 

This  result,  in  accordance  with  the  last  part  of  the  Note  to 
Art.  413,  only  expresses  the  value  of  the  fraction  for  such 
values  of  x  as  make  the  second  member  a  convergent  series. 

2.    Expand  —  — «  into  a  series. 


l-2ic-a;'2 


Assume 


1  —  3  x- 


■x* 


A  +  Bx  +  Cx2+Dx!i  +  JExi  + 


l—2x—x 

Clearing  of  fractions,  and  collecting  terms, 
1-Sx 


x2  =  A  +    B 
-2  A 


x+  C 
-2B 
-    A 


x2+  D 
-2(7 
-     B 


x3  +    E 
-2D 

-     C 


.v 


Equating  the  coefficients  of  like  powers  of  x, 
A  =  l. 
B-2A  =  -3;  whence,  B  —  2A  —  3  =  —  1. 
C-2B-A  =  -1;  whence,  C=2B  +  A-l  =  -2. 
D-2C—B=     0;  whence,  D=2  C+B  =  —  5. 
E-2D-  0—     0;  whence,  E=2D+  C=-12. 

Substituting  these  values, 


1-3 


x 


■x 


\-2x 


■X" 


r  =  1  —  x  —  2  x2  —  5  x3  —  12  x*  — ,  A  ns. 


Note.  This  method  enables  us  to  find  the  law  of  the  coefficients  in  any 
expansion.  For  instance,  in  Example  1,  we  obtained  the  equations  C=3B, 
D  =  BC,  E  =  ZD,  etc.  ;  or,  in  general,  any  coefficient,  after  the  second,  is 
three  times  the  preceding.  In  Example  2,  we  obtained  the  equations 
Ij  —  IC-^B,  E=2D  +  C,  etc.;  or,  in  general,  any  coefficient,  after  the 
third,  is  twice  the  preceding  plus  the  next  but  one  preceding.  After  the 
law  of  the  coefficients  of  any  expansion  has  been  found,  we  may  write  out 
the  subsequent  terms  to  any  desired  extent  by  its  aid. 


2-3 

x  +  Ax2 

1  +  2 

x  —  5x2' 

3  +  x 

-2x2 

3  —  x 

+  x2  ' 

1- 

-3x2 

UNDETERMINED  COEFFICIENTS.  309 

EXAMPLES. 

Expand  the  following  to  five  terms : 

3.  *  =  ".  6.  \~'-*.  9. 

1  +  X  1  +  x  +  xz 

3+1*  7         l-^2  1Q 

l-5x'  l  +  2x-3a;2' 

2  -  a;  +  x2  8       l  +  2.x  n 
Sl       1-x2     '          '   2-cc-a;2'  "    2 -3  a: -2  a- 

415.  If  the  lowest  power  of  x  in  the  denominator  is  higher 
than  the  lowest  power  of  x  in  the  numerator,  the  method  of  the 
preceding  article  will  he  found  inapplicable.  We  may,  how- 
ever, determine  by  actual  division  what  will  be  the  exponent 
of  x  in  the  first  term  of  the  expansion,  and  assume  the  fraction 
equal  to  a  series  whose  first  term  contains  that  power  of  x  ;  the 
exponents  afterwards  increasing  by  unity  each  term  as  usual. 

1.    Expand  ^ ;2  into  a  series. 

O  X  —  X 

Proceeding  in  the  usual  way,  we  should  assume 
1 


=A+Bx+ 


Clearing  of  fractions,        l  =  3Ax  +  (3B  —  A)x2  + 


Equating  the  coefficients  of  like  powers  of  x,  we  have  1  =  0 ; 
a  result  manifestly  absurd,  and  showing  that  the  usual  method 
is  inapplicable. 


x-1 


But,  dividing  1  by  3  x  —  x'1,  we  obtain  — —  as  the  first  term 

o 

of  the  quotient ;  hence  we  assume  the  fraction  equal  to  a  series 

whose  first  term  contains  x~l ;  next  term  x°,  or  1 ;  next  term 

x ;  etc.     Or, 

5 T  =  Ax~1  +  B  +  Cx  +  Dx2  +  Ex*+ 

OX  —  X' 


310  ALGEBRA. 

Clearing  of  fractions,  and  collecting  terms, 

x*  + 


1=3A+3B 
-     A 


x  +  3C 
-    B 


x-  +  oD 

-     C 


x3  +  3E 
-    D 


Equating  the  coefficients  of  like  powers  of  x, 

3  A  =  1 ;  whence,  A  =  - . 

o 

3B~A  =  0-  whence,  B  =  —  =  - . 

3       9 

7?       1 
3  C—B  —  0]  whence,  C  =  —  =  — . 

3D-  C=0;  whence, D  =  ^-  =  —  . 

3E-D  =  0;  whence,  E=^  =  ~. 

Substituting  these  values, 

1  1.11  1.1, 


3x-x2      3  9      27      '81       '243 

EXAMPLES. 

Expand  the  following  to  five  terms  : 

2  n        l  +  z-a2  .     l  —  2x*  —  at 


3sc2-2a;8  "   cc-2a2+3cc3  a-a  +  a:8-a-4 


APPLICATION    TO  THE    EXPANSION  OF  RADICALS   INTO   SERIES. 

416.  As  any  root  of  any  expression  consisting  of  two  or 
more  terms  can  be  obtained  by  the  method  of  Art.  247,  we 
know  that  the  development  is  possible. 


1.    Expand  \'l  +  x2  into  a  series  by  the  Theorem  of  Unde- 
termined Coefficients. 


Assume  \Jl-\-x'2  =  A  +  Bx  +  Cx*+  Dx*  +  22xi  + 


UNDETERMINED   COEFFICIENTS. 


311 


Squaring  both  members,  we  have  (Art.  230), 


l  +  x2  =  A2 


+  2AB 


x  +     B2 

+  2  AC 


.'■- 


+  2  AD 

+  2BC 


x3  + 


C2 
+  2AE 
+  2BD 


./■• 


Equating  the  coefficients  of  like  powers  of  x, 

A2  =  l;  whence,  A  =  l. 

0 


2AB  =  0;  whence, B  =  - 

B2  +  2AC=1;  whence,  C  = 


2  A 

1  -  B2  _  1 
^2 


2vlZ>  +  2i?CY  =  0;  whence,  D  = =0 

C2  +  2AB+2BD  =  0;  whence,  .E=- 


2  A 

BC 
A 

2BD+  C2 


2  A 


1 

8 


Substituting  these  values, 


yi  +  x2  =  l  +  -x2--;xi  + 


2         8 

which  may  be  verified  by  the  method  of  Art.  239. 

Note.     From  the  equation  A-  =  \,  we  may  have  A=  ±  1  ;  and  taking 

-4  =  - 1,  we  should  find  C  =  -  — ,  E=— ,  ,  so  that  the  expansion  might 

2  8 

he  as  follows  : 


2  8 


This  agrees  with  the  remark  made  after  the  rule  in  Art.  239. 

EXAMPLES. 

Expand  the  following  to  five  terms  : 

2.  Sjl  +  x.        4.    v/l-2a;  +  3x2. 

3.  v/1-2*.     5.  y/1  +  *  +  x2. 


6.  yi-aj. 


7.    yl  +  ic  +  a;2. 


[12  ALGEBRA. 


APPLICATION  TO   THE  DECOMPOSITION    OF  RATIONAL 

FRACTIONS. 

417.  When  the  denominator  of  a  fraction  can  be  resolved 

into  factors,  and  the  numerator  is  of  a  lower  degree  than  the 
denominator,  the  Theorem  of  Undetermined  Coefficients  ena- 
bles us  to  express  the  given  fraction  as  the  sum  of  two  or  more 
'partial  fractions,  whose  denominators  are  the  factors  of  the 
given  denominator. 

We  shall  consider  only  those  cases  in  which  the  factors  of 
the  denominator  are  all  of  the  first  degree. 

CASE    I. 

418.  When   the  factors  of  the  denominator  are  all  un- 
equal. 

x  +  7  . 

Let  — tt— 7z ^r  be  a  fraction,  whose  denominator  is 

(3  x  —  1)  (5  x  +  2) 

composed  of  two   unequal  first  degree  factors.     We  wish  to 

prove  that  it  can  be  decomposed   into    two  fractions,   whose 

denominators  are  3  x  —  1  and  5  x  +  2,  and  whose  numerators 

are  independent  of  x.     To  prove  this,  assume 

x  +  T  A  B 

+ 


(3  x  - 1)  (5  x  +  2)      3  x  -  1     5  x  +  2 

We  will  now  show  that  such  values,  independent  of  x,  may 
be  given  to  A  and  B,  as  will  make  the  above  equation  identi- 
cal, or  true  for  all  values  of  x.     Clearing  of  fractions, 

x  +  7  =  A  (5  x  +  2)  +  B  (3  x  -  1) 
or,  x  +  7  =  (5  A  +  3  B)  x  +  2  A  -  B, 

which  is  to  be  true  for  all  values  of  x.  Then,  by  Art.  413,  the 
coefficients  of  like  powers  of  x  in  the  two  members  must  be 
equal ;  or, 

5^+35=1 

2A-B=7 


UNDETERMINED   COEFFICIENTS.  313 

From  these  two  equations  we  obtain  A  =  2,  and  B  =  —  3. 
Hence,  the  proposed  decomposition  is  possible,  and  we  have 

x  +  7  2  -3 


(3x-l)(5x  +  2)      3x-l'5x  +  2 


2 


o 


3a;-l      5x  +  2 

This  result  may  be  readily  verified  by  finding  the  sum  of 
the  fractions. 

In  a  similar  manner  we  can  prove  that  any  fraction,  whose 
denominator  is  composed  of  unequal  first  degree  factors,  can 
be  decomposed  into  as  many  fractions  as  there  are  factors, 
having  these  factors  for  their  denominators,  and  for  their  nu- 
merators quantities  independent  of  x. 

EXAMPLES. 

1.    Decompose  —2 — j- —    — —  into  its  partial  fractions. 

The  factors  of  the  denominator  are  x  —  8  and  x  —  5  (Art.  118). 

a  .i  3  x  —  5  A  B  ,*  >. 

Assume,  then,     — — —  = -\ (J-) 

x-  —  Id  x  +  40      x  —  8      x  —  5 

Clearing  of  fractions,  and  uniting  terms, 

3  x  -  5  =  A  (x  -  5)  +  B  (x  -  8) 

19 

Putting  x  =  8,  19  =  3  A,  or  A  =         -g-- 

Putting  cc  =  5,  10=  —  3  B,  or  B=  -— g-- 

Note.    The  student  may  compare  the  above  method  of  finding  A  and 
B  with  that  used  on  page  312. 

Substituting  these  values  in  (1), 

19  10 

3^-5  "3"  3_19  10 

x1  - 13  x  +  40  ~~  x-S  +  x  —  5  ~  3  (x  -  8)  ~  3  (x  -  5) '  AnS' 


314  ALGEBRA. 

EXAMPLES. 

Decompose  the  following  into  their  partial  fractions 

5ic  — 2  3a;  +  2  n  cc 

«•   - o — r  •        ^   i — o — • 

ar  —  4  ar  —  2  a; 

x  +  9  _         2a;  — 3 

0. 


6. 

•> 
ar  — 

13  a-  +  42  ' 

7. 

17 

6  a;2 

- 13  x  -  5 

a; 

* 

x1  +  3  a; '  a;2  —  3  a;  —  4 

ft  7a'  +  9  Q 

9  +  9  x  -  4  a;2  *  *    (a;2  -  1)  (x-2)' 


CASE    II. 
419.      When  the  factors  of  the  denominator  are  all  equal. 

x2  —  11  x  +  26 

1.    Separate '-  3 —  into  its  partial  fractions. 

\x  —  o) 

If  we  attempt  to  perform   the  example  hy  the  method  of 
Case  I,  we  should  assume 

x*—l±x  +  2&         ABC 

+ o  + 


(x  —  3)3           x  —  3       x  —  3      x  —  3 
This  would  evidently  he  impossible,  as  the  sum  of  the  frac- 
tions in  the   second  member  is — ;  which,  as  A,  B, 

x  —  o 

and  C  are,  by  supposition,  independent  of  x,  cannot  be  equal 
a;2 -11  a; +  26 

tO  ; jr-r . 

(x  —  3)8 

The  method  to  be  used  in  Case  II  depends  on  the  following : 

n       .,      .,.  a  x"-1  +  b  xn~2  +  c x"-*  + +  k 

Consider  the  traction r . 

(aj  +  h)n 

Putting  x  =  y  —  h,  the  fraction  becomes 
a(y-  hy-1  +  b(i/  —  h)n-2  +  e  (y  —  A)""3  + +  k 


UNDETERMINED   COEFFICIENTS.  315 

If  the  terms  of  the  numerator  are  expanded  by  the  binomial 
theorem,  and  the  terms  containing  like  powers  of  y  collected 
together,  we  shall  have  a  fraction  of  the  form 

a,  y'1'1  +  br  yn~2  +  cx  yn~3  + +ky 


y, 


Dividing  each  term  of  the  numerator  by  yn,  we  have 
ax       bx       cx  l\ 

y     y     y  v 

Changing  back  y  to  x  +  h,  this  becomes 

«i  b,  c,  /.', 

+  ,     , '     ,+  ,     ,'      Q  + + 


x  +  h      (x+  h)2       (x  +  h)3  (x  +  h)n 

This  shows  that  the  assumed  fraction  can  be  expressed  as 
the  sum  of  n  partial  fractions,  whose  numerators  are  indepen- 
dent of  x,  and  whose  denominators  are  the  powers  of  x  +  h, 
beginning  with  the  first,  and  ending  with  the  nth. 

In  accordance  with  this,  we  assume 

a2 -11  a; +  26         A  B        ,         C 


(x-3)3       ~  x  -  3  T  (x  -  3)2  ^  (x  -  3)3 ' 
Clearing  of  fractions, 

x2-llx  +  26  =  A(x-3)-  +  B(x-3)+  C 

=  A  (x2 -6x  +  9)  +  B(x  —  B)+-C 
=  A  x2  +  (B  -  6  A)  x  +  9  A  -  3  B  +  C. 

Equating  the  coefficients  of  like  powers  of  x, 

A  =  l,  B-6A  =  -11,  and  9A-3B+  C=26 
Whence,  A  =  1,  B  =  -5,  and  C  =  2. 

Substituting  these  values, 

a;2-lla;  +  26_       1  5  2 

(x  -  3)3~~     "  ~x~^3  ~  (x  -  3)2  +  (x  -  3)3 '     US' 


2' 


316  ALGEBEA. 

EXAMPLES, 

Separate  the  following  into  their  partial  fractions : 

ar+3a:  +  3                      a;2  3  a;  — 10 

(x  + 1)3     "            '    (x-2)3*  (2a--5)s 

2  a; -13                      3  a;2 -4  18  a;2  +  12  a;  -  3 

(x-5)2'                    (cc  +  1)8'  (3  x  +  2/ 


CASE     III. 

420.     When  some  of  the  factors  of  the  denominator  are 
equal. 

1.    Separate zrrz  into  its  partial  fractions. 

1  x  (x  + l)3  l 

The  method  in  this  case  is  a  combination  of  the  methods  of 
Cases  I  and  II.     We  assume 

3  a +  2  A  B  C  D 

+  ,..   .    ,,,+  ,..  ,    ^„  +  — • 


a;  (x  +  l)3      a-  +  1      (x  +  l)2      (a;  +  l)3 
Clearing  of  fractions, 
3ai  +  2  =  A  x  (x  +  l)2  +  B  x  {x  +  1)  +  Cx  +  D  (x  +  l)3 

=  (A  +  D)  x* + (2A  +  B + 3  D)  x2+  ( A  +B  +  C+  3  D)  x  +B. 

Equating  the  coefficients  of  like  powers  of  x, 

£>  =  2,  A  +  B  +  C+3  B  =  3,  2A  +  B+3B  =  0,  and 

A  +  B  =  0 

Whence,      A  =  -2,  B  =  -2,   C  =  l,  andZ>  =  2. 
Substituting  these  values, 

3a;  +  2  2  2  12, 


a;(a;  +  l)3  a;  +  1       (a?  + 1)2      (a-  +  l)3      x 

It  is  impossible  to  give  an  example  to  illustrate  every  pos- 
sible case;  but  no  difficulty  will   be  found  in  assuming  the 


UNDETERMINED   COEFFICIENTS.  317 

proper  partial  fractions,  if  attention  be  given  to  the  following 
general  case.     A  fraction  of  the  form 

X 

(x  +  a)  (x  +  b) (x  +  m)r  (x  +  n)s 

should  he  put  equal  to 

A           B  E      t        F  K 

x+a       x+b x  +  m       (x+m)'  (x  +  m)r 

L  M  R 

+  x  +  n  +  (x  +  n)2  +  +  (x  +  n)s  +  

Single  factors,  like  x  +  a  and  x  +  b,  having  single  fractions 

A  B 

like  and  -     — — ,  corresponding ;    and  repeated  factors, 

x+a        -x+b 

like   (x  +  m)r,  having  r  partial  fractions   corresponding,   ar- 
ranged as  in  Case  II. 

EXAMPLES. 

Separate  the  following  into  their  partial  fractions : 

8-3  x-x'2  15  -  7  x  +  3  x2 -  3  xz 

x(x  +  2)2  '  x3  (x  +  5) 

3  ,-r3  -  11  x2  +  13  x  -  4  6  ar  -  14  x  +  6 

x  (x  -  1)  (x - 2)2      '  (x-2)(2x-3y2' 

3a;-l  ?    5  x2  +  3  x  +  2 

x2{x  +  iy  '    x3(x  +  iy2  ' 

421.     Unless  the  numerator  is  of  a  lower  degree  than  the 
denominator,  the  preceding  methods  are  inapplicable. 

:'  .'•-  +  1  . 

For  example,  let  it  be  required  to  separate  -— ^—   —  into  its 
partial  fractions.     Proceeding  in  the  usual  way,  we  assume 
2  .r-  +1       A         B 

Ob  *jC  JC  *fc  -x. 

Clearing  of  fractions, 

2  x2  +  1  =  A  (x  - 1)  +  B  x  =  (A  +  B)  x  -  A. 


318 


ALGEBRA. 


Equating  the  coefficients  of  like  powers  of  x,  we  have  2  =  0; 
an  absurd  result,  and  showing  that  the  usual  method  is  inap- 
plicable. 

But  by  actual  division,  as  in  Art.  150,  we  have 


2z2+l  2«  +  l 

—  £  + 


x-  —  x 


x- 


X 


2  x  +  1  .  .,„.-, 

We  may  now  separate  — ^ into  its  partial  fractions  by 

the  usual  method,  obtaining 

2a; +  1  1  3 


x' —  x 


X        X 


1" 


Hence, 


2a2+l      0      2^  +  1      0      1  3 

=  2  +  —r-        =2-    -  +  -     —,Ans. 


x- 


X 


x-  —  X 


APPLICATION  TO  THE  REVERSION  OF  SERIES. 

422.     1.    Given  y  =  2x  +  x2  —  2xz  —  3x*  + ,  to  revert 

the  series,  or  to  express  x  in  terms  of  y. 

Assume  x  =  Ay  +  B if  +  G if  +  D  if  + (1) 

Substituting  in  this  the  given  value  of  y,  we  have 

x=A(2x  +  x2-2xs-3xi+...)+B(ix2  +  x4+Axs-Sxi+  ...) 
+  <7(8x3  +  12z4+...)  +  Z>(16z4+...)  + 


or, 


X: 


2Ax+     A 

x2-2A 

x3-    3  A 

+  4:B 

+  4B 

-    IB 

+  8  0 

+  12  G 
+  16Z> 

x*  + 


Equating  the  coefficients  of  like  powers  of  x, 


2  A  =  1 ;  whence,  A  =  - . 


A  +  4  B—  0  ;  whence,  B— r  =  —  - , 

4  8 


-2A  +  4.B+S  G=0;  whence,  0= 


3^ 
16" 


-3^-7^+12  C+1GZ>  =  0;  whence,  D-  — 


13 
128' 


UNDETERMINED   COEFFICIENTS. 


319 


Substituting  these  values  in  (1), 

V      if      3  y3      13  yi 

If  the  even  powers  of  x  are  wanting  in  the  given  series,  we 
may  abridge  the  operation  by  assuming  x  equal  to  a  series 
containing  only  the  odd  powers  of  y. 

Thus,  to  revert   the   series   y  =  x  —  x3  +  x5  —  x1  + ,  we 

assume  x  =  A  y  +  B  y3  +  C  y5  +  D  y"  + 

If  the  odd  powers  of  x  are  wanting  in  the  given  series,  the 
reversion  of  the  series  is  impossible  by  the  method  previously 
given.  But  by  substituting  another  letter,  say  t,  for  x~,  we 
may  revert  the  series  and  obtain  a  value  of  t,  or  of  x2,  in  terms 
of  y ;  and  by  taking  the  square  root  of  the  result,  express  x 
itself  in  terms  of  y. 

If  the  first  term  of  the  series  is  independent  of  x,  we  cannot, 
by  the  method  previously  given,  express  x  definitely  in  terms 
of  y ;  though  we  can  express  it  in  the  form  of  a  series  in  which 
y  is  the  only  unknown  quantity. 

2.   Kevert  the  series  y  =  2  +  2x  —  x2  —  x3  +  2xi+ 

We  may  write  the  series, 

y-2  =  2x-x2-x3  +  2x4+ (1) 

Assume x=A(y-2)  +  B(y-2y2+C(y-2)3+D(y-2y+...  (2) 
Substituting  in  this  the  value  of  y  —  2  given  in  (1),  we  have 

x  =  A(2x—  x2—  xs+2x*+  ...)  +  B(4:X2  +  xi— ix3—  4x*+  ...) 

+  C(Sx3-12xi+  ...)  +  D(lGx4+  ...)+  


or,  x  =  2  A  x  —    A 

+  4:B 


x2-    A 

x3+    2  A 

-42? 

-    3B 

+  SC 

-12C 

+  16B 

x*  + 


320  ALGEBRA. 

Equating  the  coefficients  of  like  powers  of  x, 

2  A  —  l;  whence,  A  =  ^ . 

—  A  +  4  B  =:Jd ;  whence,  B  =  ^ . 

o 

—  A-4:B  +  8C  =  0;  whence,  C  =  ^. 

o 

7 
2A-3B-12C+16I>  =  0]  whence,  D  =  -j-. 

Suhstituting  in  (2), 

x=l(y-2)+l(1/-2y+l(}/-2y+^(u-2y+ ,Am. 

EXAMPLES. 

Revert  the  following  series  to  four  terms : 

3.  y  =  x  +  x-  +  x3  +  xi  + 

4.  ij  =  2x  +  3x3  +  4:x5+ 5x7  + 

5.  i/  =  x  —  x3  +  x5  —  x1  + 

2  3  4 

^y»—  rj/*"  'V*^ 

\K/  *Aj  *AJ 


tAs  \Kj  \Aj 


8.   y  =  3x-2x2+3x3-4xi  + 


Note.  This  method  may  sometimes  be  used  to  find,  approximately,  the 
root  of  an  equation  of  higher  degree  than  the  second.  Thus,  to  solve  the 
equation 

we  may  put  .1  =  2/,  and  revert  the  series  ;  giving,  as  in  Ex.  1,  Art.  422, 

1  1    „      3     ,       13 

x  =  —  ii if-\ if if  +  

2  &J       16  128 

Putting  back  y=.l,  we  have 

.1      .01     .003     .0013 
^                            2        8         16         128 
=  . 05-. 00125 +  . 00019 -. 00001  + =  .04893  +  ,  Ans. 

This  method  can,  of  course,  only  be  used  when  the  series  in  the  second 
member  is  convergent. 


BINOMIAL  THEOREM.  321 

XXXIX.  — BINOMIAL  THEOREM. 

ANY  EXPONENT. 
423.     We  have  seen  (Art.  402)  that  when  n  is  a  positive 


integer, 


n(n—  1)    „      n(n  —  l)(n  —  2)    „ 
(l  +  a;)»=l  +  wa;  +  — ^r V+  — ^ '-  xz  + 


We  shall  now  prove  that  this  formula  is  true  when  n  is  a 
positive  fraction,  a  negative  integer,  or  a  negative  fraction. 

1.    Let  n  be  a  positive  fraction,  which  we  will  denote  by 

p  ... 

— ;  p  and  q  being  positive  integers. 


Now  (Art.  252),    (1  +  x)  *  =  V  (1  +  xf 


=  yi+px+ ,  (Art.  402). 

Extracting  the  ^th  root  of  this  expression  by  the  method  of 


Art.  247, 


1  +px  + 
1«  =  1 


.       p  x 

1  +  —  + 


q      j)  x 


That  is,  (l  +  a:)i"  =  l  +  — + (1) 

2.    Let  n  be  a  negative  quantity,  either  integer  or  fraction, 
which  we  will  denote  by  —  s. 

Then  (Art.  255), 

= ,  (by  Arts.  402,  and  423, 1). 

1  +  sx+ '  v  J 

From  which,  by  actual  division,  we  have 

(l  +  x)-°  =  l-sx  + (2) 


322  ALGEBRA. 

From  (1),  (2),  and  Art.  402,  we  observe  that  whether  n  is 
positive  or  negative,  integral  or  fractional,  the  form  of  the 
expansion  is 

(1  +  x)n  =  1  +  n  x  +  A  x2  +  B  x3  +  Cx*  + (3) 

x 
Writing  -  in  place  of  x,  we  have 
a 

/y*  /y>-J  /y.3  /y»4 


1  +  -)   =l  +  n-  +  A~r2  +  B-+C—i  + 

al  a  a1  a6  a4 

Multiplying  this  through  by  an,  and  remembering  that 

(x\ n     r    /      x\ i " 
1+-)    =    a\l+-j      =  (a  +  x)n,  we  have 

(a  +  x)n  =  an  +  n  a"'1  x  +  A  an~2 x-  +  Ban~sxs  + (4) 

To  find  the  values  of  A,  B,  etc.,  we  put  x  +  z  for  x  in  (3), 
and  regarding  (x  +  z)  as  one  term,  we  shall  have 

[1  +  (x  +  z)']r'  =  l  +  ii(x  +  z)  +  A(x  +  z)2  +  B(x  +  z)3  + 

=  1  +  n  x  +  A  x'1  +  B  xs  + 

+  (n  +  2Ax  +  3Bx2  + )*+ (5) 

Regarding  (1  +  x)  as  one  term,  we  shall  have,  by  (4), 

[(1  +  x)  +z~]n=  (1  +  x)n  +  n  (l  +  x)n-1z+ (6) 

Since  [1  +  (x  +  z)~\n  =  [(1  +  x)  +  z\n,  identically,  we  have 
from  (5)  and  (6), 

l  +  nx  +  Ax2+Bxz  + +  (n  +  2Ax  +  3Bx2+ )z+ 

=  (1  +  a-)n  +  w(l  +  a-)n_1  *  + 

which  is  true  for  all  values  of  z  which  make  both  members  of 
the  equation  convergent.  Hence,  by  Art.  413,  the  coefficients 
of  z  in  the  two  series  must  be  equal ;  or, 

11  (l  +  x)n-1  =  n  +  2Ax  +  3Bx2+ 


BINOMIAL  THEOREM.  323 

Multiplying  both  members  by  1  +  x, 

n(l  +  x)n  =  n+  (2A  +  n)x+(3B+2A)x2  + 

or,  by  (3), 

n  +  n2 x  +  n  Ax2  +  n  Bx3  + =  n+  (2  A  +  n)  x 

+  (3B+2A)x2  + 

•which  is  true  for  all  values  of  x  which  make  both  members  of 
the  equation  convergent ;  hence,  equating  the  coefficients  of 
like  powers  of  x, 

sfy    {fl,  j\ 

2  A  +  n  =  n2 ;  whence,  2  A  =  n2  —  n,  or  A  =  -   --= 

3B+2A  =  ?iA;  whence,  3  B=n  A-2  A  =  A  (n-2) 

„      A  (n  -  2)       n  (n  -  1)  (n  -  2) 
B=~3—  = \3 

Substituting  in  (4), 

(a  +  x)n  =  an  +  na"-1x+  V  ^\~    '  an~2 x2 


n(n-l)(n-2) 


+  — i -^ an~a  Xs  + 


which  has  tbus  been  proved  to  hold  for  all  values  of  n,  positive 

or  negative,  integral  or  fractional.  Hence,  the  Binomial  The- 
orem has  been  proved  in  its  most  general  form.  The  result, 
however,  only  expresses  the  value  of  {a  +  x)n  for  such  values 
of  x  as  make  the  second  member  convergent  (Art.  413). 

424.  When  n  is  a  positive  integer,  the  number  of  terms  in 
the  expansion  is  n  +  1  (Art.  399).  When  n  is  a  fraction  or 
negative  quantity,  the  expansion  never  terminates,  as  no  one 
of  the  quantities  n  —  1,  n  —  2,  etc.,  can  become  equal  to  zero. 
The  development  in  that  case  furnishes  an  infinite  series. 


324  ALGEBRA. 

425.  The  method  and  notes  of  Art.  403  apply  to  the  ex- 
pansion of  expressions  hy  the  Binomial  Theorem  when  the 
exponent  is  a  fractional  or  negative  quantity. 

2 

1.    Expand  (a  +  x) :!  to  five  terms. 

2 
The  exponent  of  a  in  the  first  term  of  the  expansion  is  - ,  and 

o 

decreases  hy  one  in  each  succeeding  term. 

The  exponent  of  x  in  the  second  term  of  the  expansion  is  1, 
and  increases  hy  one  in  each  succeeding  term. 

The  coefficient  of  the   first  term  is  1 ;  of  the  second  term, 

2  2 

-  ;  multiplying  the  coefficient  of  the  second  term,  -,  hy  the  ex- 

o  o 

1  2 

ponent  of  a  m  that  term,  —  - ,  and  dividing  the  product,  —  - , 

hy  the  numher  of  the  term,  2,  we  ohtain  —  -  as  the  coefficient  of 
the  third  term  ;  etc. 

2     2_i       1   _4         4    -i            7      _jo 
Kesult,    a3  +  ~a   3x—-a  ^x2+^ra   3x3  —  7r-r^a    s  xA-\- 


.^1-2 


2.    Expand  (1  +  2  a:2)-2  to  five  terms. 

(l  +  2zV2=D-  +  (2*-)]- 

=  l-2  -  2.1-= .  (2  x$)  +  3.1-4 .  (2  xfy  -  4.1-5 .  (2  a;*)8 

+  5.1-6 .  (2  a;*)4 


i 


=  1  -  2  (2  x-)  +  3  (4  a)  -  4  (8  x?)  +  5  (16  a2)  -  .... 
=  l-4a2  +  12x-32a:-  +  S0a;2  — ,  Ans. 

3.    Expand  (a-1  — 3  a:    *)    *  to  five  terms. 

(a-1-3a5~i)~*  =  [(«"1)  +  (-3  a;-*)]"* 


BINOMIAL   THEOREM.  325 

=  (a-1)"^  -  \  (co-1)-'"  (-  3  aT*)  +  "  (0~V  (-  3  x~fy 

1  40  L3  _  JL  455  —JUS  1 

-TFr(0~v(-3*  l)8+iS(«-1)  ¥(-3*  *)«- 


81  v      y        v  y    '  243 

447  _i        14    lo  140    13  _a 

=  ft3_*    :w3a;  2)+         -3  (9a;-i)__^aV(_27a;   2) 

O  9  ol 

+i-*>o- 

4         7   _i  J-"  140    13    _a     455    .lb 

=  a^+4a3a;    2  +  14a  3  cc_1  +  — —  a  3  x    2  + -— -  a  3  #~2 , 

o  o 

^l?zs. 

EXAMPLES. 

Expand  the  following  to  live  terms : 

4.  O  +  z)2  8.     .8/r— — '  12.    (m~"3-_2«2)-2. 

v         y  yl  +  a;  v  ' 

5.  (1  +  ^)-6.  9.  ^^y3-  13.  (l  +  e.r1)^. 

6.  (l-x)~%.  10.   — — .  14.    (a;4  +  4aJ)t 

c2  +  d 

7.  V^1^-  11.    (aT*-8y)*.      15-    (g-i_8y-y 

426.  The  expression  for  the  rth  term,  derived  in  Art.  404, 
holds  for  any  value  of  n,  as  it  was  deduced  from  the  expansion 
which  has  been  proved  to  hold  universally. 

1.    Find  the  7th  term  of  (1  —  cc)~3. 

Here  r  =  7,  n  =  —  ^  ;  hence,  the 
o 

1       4    _7       10       13       16 
~3,_3*~3'""3"' '   "3""    "  3   .       NR      728a;6 
j th  term  = 172.3.4.5.6 (~ x)  =  "656T ' 

Ans. 


326  ALGEBRA. 

2.    Find  the  8th  term  of  (eft  +  x~%)~z. 

Here  r  =  8,  n  =  —  3  ;  hence,  the 

-3.-4.-5.-6.  —  7.-8.-9  ,  ^   „  .  _*  , 
8thterm  = 0.3.4.5.6.7 ^-W(*   ^ 

=  —  36  arb  x    3  ,  Ans. 

EXAMPLES. 

Find  the 

3.  8th  term  of  \/ a  +  x.  7.    7th  term  of  (x'1  —  ifift. 

4.  7th  term  of  (1  +  m)-4.  8.    5th  term  of . 

(n~*  —  <ry 

5.  5th  term  of  (1  —  ar)~ *.  9.    6th  term  of  (eft  +  3  x~l)~%. 

1  -2 

6.  6th  term  of  .  10.    8th  term  of  (x3  y  —  z  3)-3. 

y'x2  +  f 

427.     To  find  any  root  of  a  number  approximately  by  the 
Binomial  Theorem. 

1.    Find  the  approximate  square  root  of  10. 

y/ 10  =  10*  =  (9  +  1)  *  =  (32  +  1)* 
Expanding  this  hy  the  Binomial  Theorem, 

(3'  +  1)*  =  (3*)*  +  \  (3a)-i  -  \  (3')-*  +  1  (3T* 
111  ^ 

—  3  _L  Z     3-1  _  Z     Q-8   i  Q-5  ^       q-7    i 

-6  +  2.6         8.3     +jg.  3     -J28-3     + 

=  3-i  _1_    J_    _t_      _JL 

+  2.3      8.33  +  16.36     128. 37+ 

=  3  +  .16667  -  .00163  +  .00026  -  .00002  + 

=  3.16228+, 


BINOMIAL   THEOREM.  327 

which  is  the  approximate  square  root  of  10  to  the  fifth  decimal 
place,  as  may  be  verified  by  evolution. 

2.    Find  the  approximate  cube  root  of  26. 

$  26  =  26*  =  (27  -1)3  =  (33  -  1)* 
Expanding  this  by  the  Binomial  Theorem, 

(33  - 1)*  =  (33)^  +  \  (33)-3*  (- 1)  -  i  (33)"*  (- 1)» 

+|[(33rl(-i)3- 


1  1  K 
Q                   Q-2                   Q-5  Q-8 

-°~r6  ~9     "si"5  ~ 


=  3 


3.32     9.35     81. 38      

=  3  -  .037037  -  .000457  -  .000009  - 
=  2.962497  +  ,  Ans. 


RULE. 

Separate  the  given  number  into  two  parts,  the  first  of  whir// 
is  the  nearest  perfect  power  of  the  same  degree  as  the  required 
root.     Expand  the  result  by  the  Binomial  Theorem. 

Note.  If  the  second  term  of  the  binomial  is  small,  the  terms  in  the 
expansion  converge  rapidly,  and  we  obtain  an  approximate  value  of  the 
required  root  by  taking  the  sum  of  a  few  terms  of  the  development.  But 
if  the  second  term  is  large,  the  terms  converge  slowly,  and  it  requires  the 
sum  of  many  terms  to  insure  a  considerable  degree  of  accuracy. 

EXAMPLES. 

Find  the  approximate  values  of  the  following  to  five  deci- 
mal places : 

3.  #31.  5.   #99.  7.  #17. 

4.  #9.  6.  #29.  8.   #78. 


328  ALGEBRA. 


XL.  —  SUMMATION  OF  INFINITE  SERIES. 

428.  The  Summation  of  a  Series  is  the  process  of  finding 
a  finite  expression  equivalent  to  the  series. 

Different  series  require  different  methods  of  summation, 
according  to  the  nature  of  the  series,  or  the  law  of  its  forma- 
tion. Methods  of  summing  arithmetical  and  geometrical  series 
have  already  been  given  (Arts.  369,  377,  and  380).  Methods 
applicable  to  other  series  will  now  be  treated. 

RECURRING   SERIES. 

429.  A  Recurring  Series  is  one  in  which  each  term,  after 
some  fixed  term,  bears  a  uniform  relation  to  a  fixed  number  of 
the  preceding  terms.     Thus 

l  +  2x  +  3x2  +  Ax3+ 

is  a  recurring  series,  in  which  each  term,  after  the  second,  is 
equal  to  the  product  of  the  preceding  term  by  2  x,  plus  the 
product  of  the  next  term  but  one  preceding  by  —  x2. 

The  sum  of  these  constant  multipliers  is  called  the  scale  of 
relation  of  the  series,  and  their  coefficients  constitute  the  scale 
of  relation  of  the  coefficients  of  the  series.     For  example,  in 

the  series  1  +  2  x  +  3  x'2  +  4  xs  + ,  the  scale  of  relation  is 

2  x  —  x2,  and  the  scale  of  relation  of  the  coefficients  is  2  —  1. 

430.  A  recurring  series  is  said  to  be  of  the  first  order 
when  each  term,  commencing  with  the  second,  depends  on  the 
one  immediately  preceding;  of  the  second  order,  when  each 
term,  commencing  with  the  third,  depends  ujion  the  tiro  im- 
mediately preceding  ;  and  so  on. 

If  the  series  is  of  the  first  order,  the  scale  of  relation  will 
consist  of  one  term  ;  if  of  the  second  order,  it  will  consist  of 
two  terms  ;  and,  in  general,  the  order  and  the  number  of  terms 
in  the  scale  of  relation  will  correspond. 

431.  To  find  the  scale  of  relation  of  the  coefficients  of  a 
recurring  series. 


SUMMATION   OF  INFINITE  SERIES.  329 

1.  If  the  series  is  of  the  first  order,  it  is  a  simple  geometri- 
cal progression,  and  the  scale  of  relation  of  the  coefficients  is 
found  by  dividing  the  coefficient  of  any  term  by  the  coefficient 
of  the  preceding  term. 

2.  If  the  series  is  of  the  second  order,  let  a,  b,  c,  d,  

represent  the  consecutive  coefficients  of  the  series,  and  p  +  q 
their  scale  of  relation.     Then, 


c  =p  b  +  q  a) 
d=p c  +  q  b  j 


(4> 


to  determine  p  and  q ;  solving,  we  obtain 

ad  —  bc         ..  c2  —  b  d 

p  = 77 ,  and  q  =  —    —  . 

ac  —  b1  ac  —  b1 

3.    If  the  series  is  of  the  third  order,  let  a,  b,  c,  d,  e,  f,  

represent  the  consecutive  coefficients  of  the  series,  and  p  +  q 
+  r  their  scale  of  relation.     Then, 

d =p c  +  q b  +  r a 

e =p d  +  q  c  +  r  b 

f=X>  e  +  q  d  +  r  c 

from  which  we  can  find  p,  q,  and  ;■. 

432.  To  ascertain  the  order  of  a  series,  we  may  first  make 
trial  of  a  scale  of  two  terms,  and  if  the  result  does  not  corre- 
spond with  the  series,  we  may  try  three  terms,  four  terms,  and 
so  on,  till  the  true  scale  of  relation  is  found.  If  we  assume 
the  series  to  be  of  too  high  an  order,  the  terms  of  the  scale 

will  take  the  form  r.  • 

433.  To  find  the  sum  of  a  recurring  series,  when  the  scale 
of  relation  of  its  coefficients  is  known. 

Let 

a  +  bx  +  cx2  +  dx3  + +jxn~3+  kxn~2+  lxn~l  + 

be  a  recurring  se"ries  of  the  second  order.     Let  S  denote  the 


330  ALGEBKA. 

sum  of  n  terms  of  the  series ;  and  let  p  +  q  be  the  scale  of  re- 
lation of  the  coefficients.     Then, 

S=  a  +  b  x  +  c  x2  +  d  x3  + +  lxn~x 

p  Sx=pax+pbx2+pcx3  + +  pkxn~1+plxn 

q  Sx2  =  q  ax2  +  q  b  Xs  + +  qjxn~1+q  k  xn+  qlxn  +  1 

Subtracting  the  last  two  equations  from  the  first, 

S—p  Sx  —  q  Sx2  =  a  +  bx  —  pax—  plxn  —  q  kxn  —  qlxn  +  1 

the  rest  of  the  terms  of  the  second  member  disappearing,  be- 
cause, since  p  +  q  is  the  scale  of  relation  of  the  coefficients, 

c  =p  b  +  q  a,  d  =p  c  +  q  b, I  =p  k  +  qj. 

Therefore  we  have 

a  +  (b—p  a)  x  —  (p  I  +  q  k)  xn  —  q  I xn  +  1 


S  = 


■px  —  q  x" 


the  formula  for  finding  the  sum  of  n  terms  of  a  recurring  series 
of  the  second  order. 

But  if  n  becomes  indefinitely  great,  and  the  series  is  con- 
vergent, then  the  limiting  values  of  the  terms  which  involve 
xn  and  xn  + !  must  become  0,  and  we  have  at  the  limit 

s=a+  (p-pa)x 
1  —  p  x  —  qx2 

the  formula  for  finding  the  sum  of  an  infinite  recurring  series 
of  the  second  order. 

If  q  =  0,  then  the  series  is  of  the  first  order,  and  conse* 
quently  b  =p  a;  then, 

1  —px 

the  formula  for  finding  the  stun  of  an  infinite  recurring  series 
of  the  first  order.     (Compare  Art.  380.) 


SUMMATION   OF   INFINITE   SERIES.  331 

In  like  manner,  we  should  obtain 

a  +  (b  —  p  a)  x  +  (c  —pb  —  qa)x2 

o  — -z o 5 (o) 

1  —p  x  —  q  x-  —  r  x6  w 

the  formula  for  the  summation  of  an  infinite  recurring  series 
of  the  third  order. 

434.  A  recurring  series,  like  other  infinite  series,  originates 
from  an  irreducible  fraction,  called  the  generating  fraction. 
The  summation  of  the  series,  therefore,  reproduces  the  frac- 
tion ;  the  operation  being,  in  fact,  the  exact  reverse  of  that  in 
Art.  414. 

435.  1.  Find  the  sum  of  l  +  2cc  +  8ar+28z3+100x4+ 

We  must  first  determine  the  scale  of  relation  of  the  coeffi- 
cients. In  accordance  with  Art.  432,  we  first  assume  the 
series  to  be  of  the  second  order.  We  have  a  =  i,  b  =  2,  c==8, 
d  =  28.  Substituting  in  the  values  of  p  and  q  derived  from 
(^4),  Art.  431,  we  have  p  =  3  and  q  =  2.  To  ascertain  if  this 
is  the  proper  scale  of  relation,  consider  the  fifth  term,  100  x4 ; 
this  should  be  3  x  times  the  preceding  term,  plus  2  x2  times 
the  next  preceding  term  but  one,  or,  84  x4  +  16  x*.  This  shows 
that  the  series  is  of  the  second  order. 

Substituting  in  (1)  the  values  of  a,  b,  p,  and  q,  we  have 

l  +  (2  — 3)a;_         1  —  x 
b ~  i-Zx-2tf  ~ l-3x-2x2' 

EXAMPLES. 

Find  the  sum  of  the  following  series : 

2.    l  +  2a-  +  3x2  +  5a;3  +  8a;4+ 

0     a      ac         a  c2    „      a  c3 

3>  b^¥x  +  !Fx"-l^x+ 

4.  4  +  9a;  +  21ar+51;r5+ 

5.  l  +  3x  +  5x2  +  7x3+ 


332  ALGEBRA. 

6.  2-a  +  2a2-5a3+10ai-17a5+ 

7.  3  +  5x  +  7  x'2  +  13.x'3  +  23  a;4  +  45  x*  + 

8.  l  +  3x  +  4:x2+7x3+llx4+ 

9.  2  +  4x-x*-3xs  +  2xi  +  ±x5+ 

DIFFERENTIAL  METHOD. 

436.  The  Differential  Method  is  the  process  of  finding  any 
term,  or  the  sum  of  any  number  of  terms,  of  a  regular  series, 
by  means  of  the  successive  differences  of  its  terms. 

437.  If,  in  any  series,  we  take  the  first  term  from  the  sec-, 
ond,  the  second  from  the  third,  the  third  from  the  fourth,  and 
so  on,  the  remainders  will  form  a  new  series  called  the  first 
order  of  differences. 

If  the  differences  be  taken  in  this  new  series  in  like  manner, 
we  obtain  a  series  called  the  second  order  of  differences  ;  and 
so  on. 

Thus,  if  the  given  series  is 

1,     8,     27,     64,     125,     216,  

the  successive  orders  of  differences  will  be  as  follows : 

1st  order,  7,     19,     37,     61,     91,  

2d  order,  12,     18,     24,     30,   

3d  order,  6,     6,     6,   

4th  order,  0,     0,    

Hence,  in  this  case  there  are  only  three  orders  of  differences. 

438.  To  find  any  term  of  a  series. 
Let  the  series  be 

al)        a2>        a3>        ai)        a5>     an1        an+l>     

Then  the  first  order  of  differences  will  be 


SUMMATION   OF   INFINITE   SERIES.  333 

the  second  order  of  differences  will  be 

a3  —  2a2  +  al,  aA  —  2a3  +  a2,  a5  —  2ai  +  a3,   , 

the  third  order  of  differences  will  be 

aA  —  3  a3  +  3  a2  —  ax ,  a5  —  3  aA  +  3  a3  —  a2 ,   , 

the  fourth  order  of  differences  will  be 

a5  —  4  aA  +  6  a&  —  4  a2  +  a1}  , 

and  so  on ;  where  each  difference,  although  a  compound  quan- 
tity, is  called  a  term. 

Let  now  dx,  d2,  ds,  dA,  represent  the  first  terms  of  the 

several  orders  of  differences.     Then, 

dl  =  a2  —  al;  whence,  a2  =  a l  +  dA . 

d2  =  a3  —  2  a2  +  ax ;  whence,  a3  —  2a2  —  al+d2^2ai  +  2d1 

—  ax  +  d2  =  aA  +  2  dA  +  d2. 

d3  =  aA  —  3  as  +  3  a2  —  ax ;  whence,  aA  =  ax  +  3  dv  +  3  d2  -f-  ds . 

dA  =  a5  —  A  aA  +  6  a3  —  Aa2+al;  whence,  a5  =  al  +  4^d1  +  6d2 
+  ±d3+dA. 

We  observe  that  the  coefficients  of  the  value  of  a2  are  the 
same  as  the  coefficients  of  the  first  power  of  a  binomial;  the 
coefficients  of  the  value  of  a3  are  the  same  as  the  coefficients 
of  the  second  power  of  a  binomial ;  and  so  on.  Assume  that 
this  law  holds  for  the  nth  term ;  that  is,  that  the  coefficients 
of  the  value  of  aH  are  the  same  as  the  coefficients  of  the  (n  —  l)th 
power  of  a  binomial ;  then, 

,        „   .       (n-l)(n-2)  . 

a*  =  «i  +  (n  -  1)  (h  + Q d2 

+  („-!)(„-  2)  <«-S)d3+ (1) 

If  the  law  holds  for  the  nth  term  in  the  given  series,  it  will 
also  hold  for  the  «th  term  in  the  first  order  of  differences ;  or, 


334  ALGEBRA. 

an+1-an  =  d1+(n-l)d2  +  ^—-^f—^d;i  + (2) 


Adding  (1)  and  (2),  we  have 

(n  -  1)  (n  -  2) 


a 


,+1  =  «!+[! +  (n-l)]dl + 


(n  - 1)  + 


+ 


(w-l)(w-2)      (w-l)(w-2)(ra-3) 

~|2-  ~\T 


2 


,/., 


=  aj  +  n  dx  H r^-  [2  +  n  —  2]  d2 

(n  -  1)  («  -  2)  ro 


,        n(n-l)   ?        »(»-l)(n-2)    .  ,„ 

=  «!  +  »!(/!+  -^ ~ <4  + ^ "^3  + (3) 


where  the  coefficients  are  the  same  as  the  coefficients  of  the  rath 
power  of  a  binomial.  Hence,  if  the  law  holds  for  the  nth  term, 
it  also  holds  for  the  {n  +  l)th  term  ;  but  we  have  shown  it  to 
hold  for  the  fifth  term,  a5 ;  hence  it  holds  for  the  sixth  term  ; 
and  so  on.  That  is,  Formula  (1)  holds  for  any  term  in  the 
series. 

When  the  differences  finally  become  0,  the  value  of  the  nth 
term  can  be  obtained  exactly ;  but,  in  other  cases,  the  result  is 
merely  an  approximate  value. 

439.     To  find  the  stem  of  any  number  of  terms  of  a  series. 

Let  the  series  be 

a,     b,     c,     d,     e,   (1) 

Let  S  denote  the  sum  of  the  first  n  terms.     Assume   the 

series 

0,  a,  a  +  b,  a  +  b  +  c,  a  +  b  +  c  +  d, (2) 

in  which  the  (n  +  l)th  term  is  obviously  equal  to  the  sum  of  n 
terms  of  the  given  scries  ;  that  is,  S  is  the  (n  +  l)th  term  of 
series  (2).     Now  the  first  order  of  differences  of  series  (2)   is 


SUMMATION  OF   INFINITE  SERIES.  335 

the  same  as  series  (1)  ;  hence,  the  second  order  of  differences 
of  series  (2)  is  the  same  as  the  first  order  of  (1)  ;  the  third 
order  of  (2)  is  the  same  as  the  second  order  of  (1) ;  and  so  on. 

Then,  letting  a',  d\,  d'2,  d's, represent  the  first  term,  and 

the  first  terms  of  the  several  orders  of  differences  of  (2),  we 

have  a'  =  0,  d\  =  a,  d'2  =  dl,d's  =  d2, where  a,  dx ,  d2, 

are  the  first  term,  and  the  first  terms  of  the  several  orders  of 
differences  of  (1).  But,  by  (3),  Art.  438,  the  (n  +  l)th  term 
of  series  (2)  will  be 

n  (n  -  1)  „        n  (n  - 1)  (n  -  2) 

In  this  put  for  a',  d\,  d'2,  d'3, their  values;  then 

n  (n  —  1)  ,       n  (n  —  1)  (re  —  2)  . 

S=na  +  — ^|2 — Ld1+-  ~" Ld,+ (3) 

440.     1.    Find  the  12th  term   of  the  series  2,  6,  12,  20, 

30,  

The  successive  orders  of  differences  will  be  as  follows  : 

1st  order,         4,     6,     8,     10,   

2d  order,  2,     2,     2,    

3d  order,  0,     0,   

Then  ax  =  2,  dY  =  4,  d2  =  2,  ds,  dA, =  0,  and  n  =  12. 

Substituting  in  (1),  Art.  438,  the  12th  term 

(12  —  1)  (12  —  2) 
=  2  +  (12-l)4+l  >)}  ;  2  =  2  +  44  +  110=156,^. 

2.    Find  the  sum  of  8  terms  of  the  series  2,  5,  10,  17,  

1st  order  of  differences,     3,     5,     7,   

2d  order  of  differences,  2,     2,   

3d  order  of  differences,  0,   

Then  a  =  2,  d,  =  3,  d,  =  2,  n  =  S. 


336  ALGEBRA. 

Substituting  these  values  in  (3),  Art.  439,  we  have 

=  16  +  84  +  112  =  212,  Ans. 

EXAMPLES. 

3.  Find  the  first  term  of  the  fifth  order  of  differences  of 
the  series  6,  9,  17,  35,  63,  99,  

4.  Find  the  first  term  of  the  sixth  order  of  differences  of 
the  series  3,  6,  11,  17,  24,  36,  50,  72,  

5.  Find  the  seventh  term  of  the  series  3,  5,  8,  12,  17,  

6.  Sum  the  first  twelve  terms  of  the  series  1,  4,  10,  20, 
35, 

7.  Sum  the  first  hundred  terms  of  the  series  1,  2,  3,  4, 
5, 

8.  Find  the  15th  term  of  the  series  l2,  22,  32,  42,  

9.  Sum  the  first  n  terms  of  the  series  l3,  23,  33,  43,  53,  

10.    Sum  the  first  n  terms  of  the  series  1,  24,  34,  44,  54,  64, 


11.  If  shot  be  piled  in  the  shape  of  a  pyramid,  with  a  trian- 
gular base,  each  side  of  which  exhibits  9  shot,  find  the  number 
contained  in  the  pile. 

12.  If  shot  be  piled  in  the  shape  of  a  pyramid,  with  a  square 
base,  each  side  of  which  exhibits  25  shot,  find  the  number 
contained  in  the  pile. 

INTERPOLATION. 

441.  Interpolation  is  the  process  of  introducing  between 
terms  of  a  series  other  terms  conforming  to  the  law  of  the 

series. 


SUMMATION   OF   INFINITE   SERIES.  337 

Its  usual  application  is  in  finding  intermediate  numbers 
between  tbose  given  in  Mathematical  Tables,  which  may  be 
regarded  as  a  series  of  equidistant  terms. 

442.  The  interpolation  of  any  intermediate  term  in  a 
series,  is  essentially  finding  the  nth  term  of  the  series,  by  the 
differential  method  (Art.  438).     Thus, 

Let  t  represent  tbe  term  to  be  interpolated  in  a  series  of 
equidistant  terms,  and  p  the  distance  the  term  t  is  removed 
from  the  first  term,  a,  expressed  in  intervals  and  fractions  of 
an  interval ;  that  is,  p  being  the  distance  to  the  nth  term, 
p  =  n  —  1  intervals. 

In  Formula  (1),  Art.  438,  putting  p  for  n  —  1,  the  nth.  term 

t=a+pdl+pj£!=v.il+p(*-1np-*>dt+ 

443.  1.    In  the  series  :j^ ,  7-7  >  TE  >  T£  >  T??  > ,  find  the 

13    14    15    16    17 

middle  term  between  5-=  and  ^-7 . 

15  lb 

Here,  the  first  differences  of  the  denominators  are 

1,     1,     1,     1,   

The  second  differences  are 

0,    0,    0,  

Whence,  dt  =  1,  and  d2  =  0. 

5 
The  distance  to  the  required  term  is  2\  intervals,  or  p  =  ^- 

Make  a  =  13,  the  denominator  of  the  first  term  ;  then  by  the 
preceding  formula,  the  denominator  of  the  required  term, 

1        2 

Therefore  the  required  term  is  —  or  757,  Ans. 

31       ol 

~2 


338  ALGEBRA. 

2.    Given  ^94  =  9.69536,  ^95  =  9.74679,  y/  96  =  9.79796  ; 
to  find  y/94i. 

Here,  the  first  differences  are 

.05143,     .05117,   

and  the  second  differences  are 

-.00026,    

Whence,  ^  =  .05143,  d2  =  -. 00026,   


.    1  .  1 

The  distance  of  the  required  term  is  -  interval,  or  p  =  -? , 

Then  the  required  term, 

1(|-1 
*  =  9.69536  +  1  x  .05143  +  4  ^  (-  ,0( 

=  9.69536  +  .01286  -  J,  (-  .00026)  + 

=  9.69536  +  .01286  +  .00002  + 

=  (approximately)  9.70824,  Ans. 


EXAMPLES. 

3.  Given  ^64  =  4,  ^65  =  4.0207,  f  66  =  4.0412,  ^67  = 
4.0615  ;  find  ^66A 

4.  Given  ^45  =  3.556893,  ^47  =  3.608826,  ^49  =  3.659306, 
^51  =  3.708430;  find  ^48. 

5.  Given  ^5  =  2.23607,  yf  6  =  2.44949,  ^7  =  2.64575,  y/8 
=  2.82843  ;  find  \'r>M. 

6.  (Jivcn  the  length  of  a  degree  of  longitude  in  latitude 
41°=4528  miles;  in  latitude  42°  =  44.59  miles;  in  latitude 
43°=43.8S  miles;  in  latitude  44°  =  43.16  miles.  Find  the 
length  of  a  degree  of  longitude  in  latitude  11'  30'. 


LOGARITHMS.  339 

7.  If  the  amount  of  $  1  at  7  per  cent  compound  interest  for 
2  years  is  $  1.145,  for  3  years  $  1.225,  for  4  years  $  1.311,  and 
for  5  years  $  1.403,  what  is  the  amount  for  4  years  and  6 
months  ? 


XLI.  —  LOGARITHMS. 

444.  The  logarithm  of  a  quantity  to  any  given  base,  is  the 
exponent  of  the  power  to  which  the  base  must  be  raised  to  equal 
the  quantity. 

For  example,  if  ax  =  m,  x  is  the  exponent  of  the  power  to 
which  the  hase,  a,  must  be  raised  to  equal  the  quantity,  m; 
or,  x  is  the  logarithm  of  m  to  the  base  a ;  which  is  briefly 
expressed  thus  : 

x  =  loga  m. 

445.  If  a  remain  fixed,  and  m  receive  different  values,  a 
certain  value  of  x  Avill  correspond  to  each  value  of  m;  and 
these  values  of  x  taken  together  constitute  a  System  of  Loga- 
rithms. And  as  the  base,  «.,  may  have  any  value  whatever,  the 
number  of  possible  systems  is  unlimited. 

For  example,  suppose  a  =  3. 

Then,  since  3°  =  1,  by  Art.  444,  0  =  log.,  1 

"  3X  =  3,      "         "       l  =  log33 

"  32=9,      "         "       2  =  log3  9 

Hence,  in  the  system  whose  base  is  3,  log  1  =  0,  log  3  =  1, 
log  9  =  2,  etc. 

Again,  suppose  a  =  12. 

Then,  since  121  =  12,     1  =  log12  12 

"  122  =  144,  2  =  log12 144 

Hence,  in  the  system  whose  base  is  12,  log  12  =  1,  log 
144  =  2,  etc. 


340  ALGEBEA. 

446.  The  only  system  in  extensive  use  for  numerical  com- 
putations is  the  Common  System  or  Briggs'  System,  whose 
base  is  10.  Therefore  the  definition  of  the  common  logarithm 
of  a  quantity  is  the  exponent  of  that  power  of  1.0  which  equals 
the  quantity.     Hence, 

Since     10°  =  1,  log10 1  =  0 

101  =  10,  log10 10  =  1 

"        102=100,  log10100  =  2 

TO3  =  1000,  log101000  =  3 

"     10-1  =  i  =  .l,  login.l  =  -l 


a 


10-2  =  ^  =  .01,  log10.01=-2 

'<     10-3-^  =  .001,         log10.001  =  -3,  etc. 

447.  It  is  customary  in  using  common  logarithms  to  omit 
the  subscript  10  which  denotes  the  base ;  hence,  we  may  write 
the  results  of  Art.  446  as  follows  : 

log  1  =  0  log  .1=- 1  =  9- 10 

log  10  =  1  log  .01  =  -  2  =  8  -  10 

log  100  =  2  log  .001  =  -3  =  7 -10 

log  1000  =  3  etc. 

The  second  form  of  the  results  in  the  second  column  will  be 
found  less  complicated  in  the  solution  of  examples. 

448.  We  infer  the  following  from  the  first  column  of 
Art.  447 : 

The  logarithm  of  any  number  between  1  and  10,  lies  between 
0  and  1. 

The  logarithm  of  any  number  between  10  and  100,  lies  be- 
tween 1  and  2. 


LOGARITHMS.  341 

The  logarithm  of  any  number  between  100  and  1000,  lies  be- 
tween 2  and  3,  etc. 

Or,  in  other  words, 

The  logarithm  of  any  number  with  one  figure  to  the  left  of 
its  decimal  point,  is  equal  to  0  plus  some  decimal. 

The  logarithm  of  any  number  with  two  figures  to  the  left  of 
its  decimal  point,  is  equal  to  1  plus  some  decimal. 

The  logarithm  of  any  number  with  three  figures  to  the  left 
of  its  decimal  point,  is  equal  to  2  plus  some  decimal,  etc. 

449.  Reasoning  in  the  same  way  from  the  second  column 
of  Art.  447, 

The  logarithm  of  any  number  between  1  and  .1,  lies  between 
0  and  9  —  10,  or  between  10  —  10  and  9  —  10. 

The  logarithm  of  any  number  between  .1  and  .01,  lies  be- 
tween 9  —  10  and  8  —  10. 

The  logarithm  of  any  number  between  .01  and  .001,  lies  be- 
tween 8  —  10  and  7  —  10,  etc. 

Or,  in  other  words, 

The  logarithm  of  any  decimal  with  no  zeros  between  its 
point  and  first  figure,  is  equal  to  9  plus  some  decimal  — 10. 

The  logarithm  of  any  decimal  with  one  zero  between  its 
point  and  first  figure,  is  equal  to  8  plus  some  decimal  — 10. 

The  logarithm  of  any  decimal  with  two  zeros  between  its 
point  and  first  figure,  is  equal  to  7  plus  some  decimal  — 10, 
etc. 

450.  It  will  be  seen  from  the  two  preceding  articles  that 
in  general  the  logarithm  of  a  number  consists  of  two  parts, 
one  integral,  the  other  decimal.  The  integral  part  is  called 
the  characteristic ;  the  decimal  part,  the  mantissa.  For  rea- 
sons which  will  be  given  hereafter,  only  the  mantissa  of  the 
logarithm  is  given  in  the  tables ;  the  characteristic  must  be 
supplied  by  the  reader.  The  rules  for  characteristic  are  based 
on  the  results  obtained  in  the  last  parts  of  Arts.  448  and  449. 


342  ALGEBRA. 

451.  I.  If  the  number  is  greater  than  1,  the  characteristic 
is  1  less   than  the  number  of  figures  to  the  left  of  the  decimal 

[lit';  lit. 

For  example,  characteristic  of  log  354.89  =  2, 

characteristic  of  log  906328.3  =  5,  etc. 

II.   If  fin'  number  is  less  than  1,  the  characteristic  is  found 

by  subtracting  the  number  of  zeros  between  the  decimal  point 
cut/  first  significant  figure  from  9;  writing  —  10  after  the 
mantissa. 

For  example,  characteristic  of  log  .00792  =  7,  with  — 10 
after  the  mantissa;  characteristic  of  log  .2583  =  9,  with  —10 
after  the  mantissa ;  etc. 

It  is  customary  in  ordinary  computation  to  omit  the  — 10 
after  the  mantissa ;  it  should  he  remembered,  however,  that  it 
is  really  a  part  of  the  logarithm,  and  should  be  allowed  for, 
and  subjected  to  precisely  the  same  operations  as  the  rest  of 
the  logarithm.  Beginners  will  find  it  useful  to  write  it  in 
all  cases ;  and  in  some  problems  it  cannot  conveniently  be 
omitted. 

Note.  Many  writers,  in  dealing  with  the  characteristics  of  the  loga- 
rithms of  numbers  less  than  1,  combine  the  two  portions  of  the  characteris- 
tic, writing  the  result  as  a  negative  characteristic  before  the  mantissa. 
Thus,  instead  of  such  an  expression  as  7.603582-10,  the  student  will  fre- 
quently find  3.6035S2  ;  a  minus  sign  being  written  over  the  characteristic, 
to  denote  that  it  alone  is  negative,  the  mantissa  being  always  positive.  The 
objection  to  this  notation  is  the  inconvenience  of  using  numbers  partly 
positive  and  partly  negative. 

PROPERTIES   OF  LOGARITHMS. 

452.  In  any  system  the  logarithm  of  unity  is  zero. 
For,  since  a0  =  1,  for  any  value  of  a,  0  =  loga  1. 

453.  In  any  system  the  logarithm  of  the  base  itself  is 
unity. 

For,  since  a1  =  a,  for  any  value  of  a,  1  =  log„  a. 


LOGARITHMS.  343 

454.  hi  any  system,  whose  base  is  greater  than  unity,  the 
logarithm  of  zero  is  minus  infinity. 

For,  since  or00  =  — -  =  —  =  0,  —  cc  =  loga  0. 
a         go 

If  the  base  is  less  than  unity,  the  logarithm  of  0  is  +  cc  . 

455.  In  any  system  the  logarithm  of  the  product  of  any 
number  of  factors  is  equal  to  the  sum  of  the  logarithms  of 
those  factors. 

Assume  the  equations, 


-  "H  whence,  by  Art.  444,  [x  ~~  !°g" 


Multiplying,         ax  X  cC>  —  m  n,  or  ax  +  \  —  m  n 

Whence,  x  +  y  =  loga  m  n 

Substituting  values  of  x  and  y, 

loga  m  n  —  loga  m  +  loga  n. 

If  there  are  three  factors,  m,  n,  andj?, 

loga  m  np  =  loga  (m  n  Xp)  =  (Art.  455)  loga  m  n  +  \ogap 

=  loga  m  +  loga  n  +  logap. 

An  extension  of  this 'method  will  prove  the  theorem  for  any 
number  of  factors. 

By  the  application  of  this  theorem,  we  may  find  the  loga- 
rithm of  a  number,  provided  we  know  the  logarithm  of  each 
of  its  factors.  For  example,  given  log  2  =  0.301030,  log  3  = 
0.477121,  required  log  72. 

log  72  =  log  (2  x  2  x  2  x  3  X  3) 

=  log  2  +  log  2  +  log  2  +  log  3  +  log  3 

=  3  x  log  2  +  2  x  log  3 

=  0.903090  +  0.954242  =  1.857332,  Ans. 


344  ALGEBRA. 

EXAMPLES. 

Given  log  2  =  0.301030,  log  3  =  0.477121,  log  7  =  0.845098, 
calculate  : 

• 

1.  log  48.  4.   log  98.  7.   log  1G8.  10.   log  3087. 

2.  log  441.         5.   log  84.  8.   log  7056.        11.   log  15552. 

3.  log  56.  6.   log  567.        9.   log  504.  12.   log  14406. 

456.  In  any  system  the  logarithm  of  a  fraction  is  equal 
to  the  logarithm  of  the  numerator  minus  the  logarithm  of  the 
denominator. 

Assume  the  equations, 


ax  =  m)     i  f  x  =  loga  m 

„  >  whence,  {         ,  oa 

ay  —  n  J  (y  =  \ogan 


Dividing, 

ar      m                ,       m 
—  =  — ,  or  ax~v  =  — 
a^       n                        n 

Whence, 

x-y  =  \oga-- 

Substituting 

values  of  x  and  y, 

loga  —  =  loga  m  —  loga  n, 

IV 

By  this  theorem,  a  logarithm  being  given,  we  may  derive 
certain  others  from  it.  For  instance,  if  we  -know  log  2  = 
0.301030,  then 

log  5  =  log  ^  =  i0g  io  -  log  2  =  1.  -  0.301030  =  0.698970. 

41 


EXAMPLES. 

Given  log  2  =  0.301030,  log  3  =  0.477121,  log  7  =  0.845098, 
calculate  : 


LOGARITHMS.  345 


1.  log  15.  4.   log  175.  7.  logTf 

2.  log  125.  5.  log  3i.  8.   log—, 


10 

T 


3.   log—.  6.  loglH.  9.   log5£. 


457.  In  any  system  the  logarithm  of  any  power  of  a 
quantity  is  equal  to  the  logarithm  of  the  quantity,  multiplied 
by  the  exponent  of  the  power. 

Assume  the  equation, 

ax  =  in,  whence,  x  =  loga  m 

Eaising  both  members  of  the  assumed  equation  to  the^th 

power, 

(ax)P  =  mP,  or  aPx  =  mP 

Whence,  px  =  log„  mP 

Substituting  the  value  of  x, 

logamP=p\ogam. 

458.  In  any  system  the  logarithm  of  any  root  of  a  quan- 
tity is  equal  to  the  logarithm  of  the  quantity,  divided  by  the 
index  of  the  root. 

For,  loga  v'  m  =  loga  (m7)  =  (Art.  457)  -  loga  m. 

459.  In  the  common  system,  the  mantissa;  of  the  loga- 
rithms of  all  numbers  having  the  same  sequence  of  figures 
will  be  the  same. 

For  example,  suppose  we  know  that  log  3.053  =  .484727. 

Then,log30.53=log(3.053xl0)=log3.053+logl0=.484727 
+  1  =  1.484727. 

Also,  log  30530  =  log  (3.053  x  10000)  =  log  3.053  +  log  10000 
=  .484727  +  4  =  4.484727. 


346  ALGEBRA. 

Again,  log.03053=log  (^~J=log3.053-logl00=.484727 

-2=  .484727  +  8-10  =  8.484727-10. 

It  is  clear,  then,  that  if  a  number  he  multiplied  or  divided  hy 
any  integral  power  of  10,  thereby  producing  another  number 
having  the  same  sequence  of  figures,  the  mantissse  of  their 
logarithms  will  be  the  same. 

Or,  to  illustrate,  if  log  3.053  =  .484727, 

then,     log  30.53  =  1.484727  log  .3053  =  9.48472?  - 10 

log  305.3  =  2.484727  log  .03053  =  8.484727  -  10 

log  3053.  =  3.484727  log  .003053  =  7.484727  -  10 

etc.  etc. 

We  may  now  see  the  reason  why,  as  stated  in  Art.  450,  only 
the  mantissa?  are  given  in  the  table ;  for  if  we  wish  to  find  the 
logarithm  of  any  number,  we  have  only  to  find  the  mantissa 
of  the  sequence  of  figures  composing  it  from  the  table,  and  can 
prefix  the  proper  characteristic,  depending  on  the  position  of 
the  decimal  point,  in  accordance  with  the  rules  stated  in  Art. 
451.  This  property  of  logarithms  is  only  enjoyed  by  the  com- 
mon system,  and  constitutes  its  superiority  over  all  others. 

460.  Given  the  logarithm,  of  a  quantity  to  a  certain  base, 
to  calculate  the  logarithm  of  the  same  quantity  to  any  other 
base. 

Assume  the  equations, 

ax  =  m)     i  (x  =  log,  m 

,,.  >  whence,  <         ,  oa 

O'J  =m)  '  \y  =  log6  m 

From  the  assumed  equations,   ax  =  by 

l  I  £ 

Hence,  (a*)?/  =  (tity,  or  av  =  b 

Whence,  -  =  lo£„  b 

y 

X 

°h  y 


l°ga  h 


LOGARITHMS.  347 

Substituting  the  values  of  x  and  y, 

,  log„  m 

log,,  m  =  .  . 

log„£ 

That  is,  if  we  know  the  logarithm  of  m  to  a  certain  base,  a, 

its  logarithm  to  any  other  base,  b,  is  found  by  dividing  by  the 

logarithm  of  b  to  the  base  a. 

461.  To  show  that  loga6  X  log6a  =  l,  for  any  values  oj 
a  and  b. 

Assume  the  equation, 

ax  =  b,  whence  x  =  loga  b 

Taking  the  -  power  of  both  members, 

00 

III 
(axy  =  bx,  or  bx  =  a 

Whence,  -  =  log6  a 

7  x 

Therefore,         loga  b  X  log6  a  =  x  X  -  =  1- 

462.  We  append  a  few  examples  to  illustrate  the  applica- 
tions of  Arts.  455,  456,  457,  and  458. 

e 

L  l0g©d=4l0g!'  (Art.  457) 

=  -  (log  a  -  log  b),    (Art.  45G) . 

Co 

2.  log  %a*Vb  =  log  Q  a  x  m\J  b)  -  log  %  e,  (Art.  456) 

=  log  \f  a  +  log  y7^  —  log  y/  c,         (Art.  455) 

=  -  log  a-\ log  b log  c,      (Art.  458). 

n  m  P 


The  following  are  proposed  as  exercises 

a  b  c 
~d~e. 


3.  log  -7—    =  log  a  +  log  b  +  log  c  —  log  d  —  log  e. 


348  ALGKBBA. 

4.  log  (J/«xJ3X  c-)  =  -  log  «  +  3  log  &  +  ^  log  a. 

5.  log^  =  r,  log2--log3. 

3g      o  o 

6.  log  \7  —  =  -  (2  log  a  —  log  b  —  log  c). 

V  0  c      n 

„    .      V  a  b      1  „  _      _ .        1 

7-  log  -^y—  =  -  (log  «  +  log  &)  -  —  log  c. 

8-  l°g      >      =  t  log  a  —  log  &  —  s  log  c  —  2  log  tf. 

bc$d2     4  ° 

(  s  I  a  _  ™  \      1  to 

9-  loS  VV  ^  +  (c  d)    "J  =  5  (loS  «  -  log  ^)  +  —  (logc+  logd)- 


USE  OF  THE  TABLE. 

463.  The  table  (Appendix)  gives  the  mantissa?  of  the 
logarithms  of  all  numbers  from  1  to  10000,  calculated  to  six 
decimal  places.  On  the  first  page  of  the  table  are  the  loga- 
rithms of  the  numbers  between  1  and  100.  This  table  is 
added  simply  for  convenience,  as  the  same  mantissae  are  to  be 
found  in  the  rest  of  the  table. 

To  find  the  logarithm  of  any  member  consisting  of  four 
figures. 

Find,  in  the  column  headed  N,  the  first  three  figures  of  the 
given  number.  Then  the  mantissa,  (if  the  required  logarithm 
will  be  found  in  the  horizontal  line  corresponding,  in  the  ver- 
tical column  which  lias  the  fourth  figure  of  the  given  number 
at  the  top.  If  only  the  last  four  figures  of  the  mantissa  are 
found,  the  first  two  figures  may  be  obtained  from  the  nearest 
mantissa  above,  in  the  same  vertical  column,  which  consists  of 
six  figures.  Finally,  prefix  the  proper  characteristic  (Art. 
451). 


LOGARITHMS.  349 

For  example,  log  140.8  =  2.148603 

log  .05837  =  8.766190  - 10 
log  8516.  =  3.930236 

For  a  number  consisting  of  one  or  two  figures,  use  the  first 
page  of  the  table,  which  needs  no  explanation  ;  for  a  number 
of  three  figures,  look  in  the  column  headed  1ST,  and  take  the 
mantissa  corresponding  in  tbe  column  headed  0.  For  exam- 
ple, log  94.6  =  1.975891. 

464.  To  find  the  logarithm  of  a  number  of  more  than  four 
figures. 

For  example,  let  it  be  required  to  find  log  3296.78. 

From  the  table,  we  find  log  3296  =  3.517987 

log  3297  =  3.518119 

That  is,  an  increase  of  one  unit  in  the  number  produces  an 
increase  of  .000132  in  the  logarithm.  Then'  evidently  an  in- 
crease of  .78  unit  in  the  number  will  produce  an  increase  of 
.78  X  .000132  in  the  logarithm  =  .000103  to  the  nearest  sixth 
decimal  place; 

Therefore,  log  3296.78  =  log  3296  +  .000103 

=  3.517987  +  .000103  =  3.518090,  Ans. 

Note.  The  foregoing  method  is  based  upon  the  assumption  that  the 
differences  of  logarithms  are  proportional  to  the  differences  of  their  corre- 
sponding numbers,  which  is  not  strictly  correct,  but  is  sufficiently  exact 
for  practical  purposes. 

We  derive  the  following  rule  from  the  above  operation : 

Find  in  the  table  the  mantissa  of  the  first  four  figures, 
without  regard  to  tin'  position  of  the  decimal  point. 

Find  the  difference  between  this  and  the  mantissa  of  the 
next  7iigher  number  of  four  figures  ;  (called  the  tabular  dif- 
ference, and  to  be  found  in  the  column  headed  D  on  each 
page  :  see  Note  on  page  350.) 

Multiply  the  tubular  difference  by  the  rest  of  the  figures  of 
the  given  number,  with  a  decimal  point  before  them. 

Add  the  result  to  the  mantissa  of  the  first  four  figures. 

Prefix  the  proper  characteristic. 


350  ALGEBRA. 

1.  Find  the  logarithm  of  .02243076. 

Mantissa  of  2243  =  350829 

Tabular  difference  =  194  lj> 

.076  350844 


1.164 
13.58 


Correction  =  14.744  =  15  nearly. 

Am.  8.350844-10. 

Note.  To  find  the  tabular  difference  mentally,  subtract  the  last  figure 
of  the  mantissa  from  the  last  figure  of  the  next  larger,  and  take  the  nean  si 
whole  number  ending  in  that  figure  to  the  number  in  the  column  headed 
D  in  the  same  line.  For  instance,  in  finding  log  .02243076,  the  last  figure 
of  the  mantissa  of  2243  is  9,  and  of  the  next  larger  mantissa,  3  ;  9  from  13 
leaves  4,  and  the  nearest  number  ending  in  4  to  193,  the  number  in  the 
column  headed  D,  is  194,  the  proper  tabular  difference. 

EXAMPLES. 

Find  the  logarithms  of  the  following  numbers : 


10  <"8oa""1 

.110    \ji. 

"'^     ^""tTlMg      XX 

LLJ_U.k/^>Xi 

z>  ■ 

2. 

.053. 

6. 

33.0908. 

10. 

912.2:..".. 

3. 

51.8. 

7. 

.0002851. 

11. 

.870092. 

4. 

.2956. 

8. 

65000.63. 

12. 

7303. Oi  8 

5. 

1.0274. 

9. 

.001030741. 

13. 

.0436927 

14.  Given  log  7.83  =  .89376,  log  7.84  =  .89432;   find  log 
78309. 

15.  Given  log  .05229  =  8.718419-10,  log  .05230  =  8.718502 
-10;  find  log  52.2938. 

16    Given  log  315.08  =  2.4984208,  log  315.09 =2.4984346; 
find  log  .003150823. 

17.  Given  log  18.84  =  1.275081,  log  18.87  =  1.275772 ;  find 
log  .188527. 

18.  Given  log  9.5338  =  .9792660,  log  9.5342  =  .9792843  ; 
find  log  9534071. 


LOGARITHMS.  351 

465.     To  find  the  number  corresponding  to  a  logarithm. 

For  example,  let  it  be  required  to  find  the  number  whose 
logarithm  is  3.693845. 

Since  the  characteristic  depends  only  on  the  position  of  the 
decimal  point,  and  in  no  way  affects  the  sequence  of  figures 
corresponding,  we  ought  to  obtain  all  of  the  number  corre- 
sponding, except  the  decimal  point,  by  considering  the  man- 
tissa only.  We  find  in  the  table  the  mantissa  693815,  of  which 
the  corresponding  number  is  4941,  and  the  mantissa  693903, 
of  which  the  corresponding  number  is  4(.)42. 

Tbat  is,  an  increase  of  88  in  the  mantissa  produces  an  in- 
crease of  one  unit  in  the  number  corresponding.  Hence,  an 
increase  of  30  in  the  mantissa  will  produce  an  increase  of  §§  of 
a  unit  in  the  number,  or  .34  nearly.     Therefore, 

Number  corresponding  =  4941  +  -34  =  4941.34,  Ans. 

We  base  the  following  rule  on  the  above  operation  : 

Find  in  the  table  the  next  less  mantissa,  the  four  figures 
corresponding,  and  the  tabular  difference. 

Subtract  the  next  less  mantissa  from  the  given,  mantissa. 

Divide  the  remainder  bg  the  tabular  difference /  (the  quo- 
tient in  general  cannot  be  depended  upon  to  more  than  two 
decimal  places.) 

Annex  all  of  the  quotient  except  the  decimal  point  to  the 
first  four  figures  of  the  number. 

Point  off. 

Note.  The  rules  for  pointing  off  are  the  reverse  of  tire  rules  for  charac- 
teristic given  in  Art.  451  : 

I.  If  — 10  is  not  written  after  the  mantissa,  add  1  to  the 
characteristic,  giving  the  number  of  figures  to  the  left  of  the 
decimal  point. 

II.  If — 10  is  written  after  the  mantissa,  subtract  the 
characteynstic  from  9  ;  giving  the  number  of  zeros  to  be 
placed  between  the  decimal  point  and  first  figure. 


352  ALGEBRA. 

1.    Find  the  number  whose  logarithm  is  7.950185  — 10. 

950185 
Next  less  mantissa =950170;  four  figures  corresponding =8916. 

Tabular  difference =49)  15.00  (.31  nearly. 

147 

~30 

Therefore,  number  corresponding  =  .00891631,  Ans. 

EXAMPLES. 

Find  the  numbers  corresponding  to  the  following : 

2.  1.880814.  6.  8.044891-10.        10.  0.990191. 

3.  9.470410-10.        7.  2.270293.  11.  7.115658-10. 

4.  0.820204.  8.  9350064-10.        12.  8.535003-10. 

5.  4.745126.  9.  3.000027.  13.  1.670180. 

14.  Given  log  113  =  2.05308,  log  114  =  2.05690 ;  find  num- 
ber corresponding  to  1.05411. 

15.  Given  log  .08630  =  8.936011  - 10,  log  .08631  =  8.936061 
—  10  ;  find  number  corresponding  to  0.936049. 

16.  Given  log  2.0702  =  .3160123,  log  2.0703  =  .3160333  ; 
find  number  corresponding  to  9.3160138  —  10. 

17.  Given  log  548  3  =  2.739018,  log  548.9  =  2.739493  ;  find 
number  corresponding  to  7.739416  —  10. 

18.  Given  log  7.3488  =  .8662164,  log  7.3492  =  .8662401 ; 
find  number  corresponding  to  2.8662350. 

466.  In  the  application  of  Arts.  455,  456,  457,  and  458,  we 
have  to  perform  the  operations  of  Addition,  Subtraction,  Mul- 
tiplication, and  Division  with  logarithms.  As  some  of  the 
problems  which  may  arise  arc  peculiar,  wo  give  a  few  hints  as 
to  their  solution,  which  will  be  found  of  service. 

1.  Addition.  If,  in  the  sum,  — 10, —20, —30,  etc.,  are 
written  after  the  mantissa,  and  the  characteristic  standing  be- 


LOGARITHMS.  353 

fore  the  mantissa  is  greater  than  9,  subtract  from  both  parts 
of  the  logarithm  such  a  multiple  of  10  as  will  make  the  charac- 
teristic before  the  mantissa  less  than  10. 

For  example,  13.354802  - 10  should  be  changed  to  3.354802 ; 
28.964316  -  30  should  be  changed  to  8.964316  -  10  ;  etc. 

2.  Subtraction.  In  subtracting  a  larger  logarithm  from 
a  smaller,  or  in  subtracting  a  negative  logarithm  from  a  posi- 
tive, the  characteristic  of  the  minuend  should  be  increased  by 
10,  — 10  being  written  after  the  mantissa  to  compensate. 

For  example,  to  subtract  3.121468  from  2.503964,  we  write 
the  minuend  in  the  form  12.503964  — 10 ;  subtracting  from 
this  3.121468,  we  have  as  a  result  9.382496  —  10. 

To  subtract  9.635321  —  10  from  9.583427  -  10,  we  write 
the  minuend  in  the  form  19.583427  —  20 ;  subtracting  from 
this  9.635321  -  10,  we  have  as  a  result  9.948106  - 10. 

3.  Multiplication.  The  hint  already  given  for  reducing 
the  result  of  Addition,  applies  with  equal  force  to  Multiplication. 

To  multiply  a  logarithm  by  a  fraction,  multiply  first  b}r  the 
numerator,  and  divide  the  result  by  the  denominator. 

4.  Division.  In  dividing  a  negative  logarithm,  add  to 
both  parts  of  the  logarithm  such  a  multiple  of  10  as  will  make 
the  quantity  after  the  mantissa  exactly  divisible  by  the  divisor, 
with  —  10  as  the  quotient. 

For  example,  to  divide  7.402938  —  10  by  6,  we  add  50  to 
both  parts  of  the  logarithm,  giving  57.402938  —  60.  Dividing 
this  by  6,  we  have  as  a  result  9.567156  — 10. 

EXAMPLES. 

1.  Add  9.096004  -  10,  4.581726,  and  8.447510  -  10. 

2.  Add  7.196070  - 10,  8.822209  -  10,  and  2.205683. 

3.  Subtract  0.659321  from  0.511490. 

4.  Subtract  7.901338  -  10  from  1.009800; 

5.  Subtract  9.156243  -  10  from  8.750404  -  10. 


354  ALGEBRA. 


6.  Multiply  9.105107  - 10  by  3. 

7.  Divide  8.452G33  - 10  by  4. 

8.  Divide  9.670392  - 10  by  11. 

9.  Multiply  9.6G8311  -  10  by  ?. 


SOLUTIONS   OF  ARITHMETICAL   PROBLEMS   BY 

LOGARITHMS. 

467.  In  finding  the  value  of  any  arithmetical  quantity  by 
logarithms,  we  first  find  the  logarithm  of  the  quantity,  as  in 
Art.  462,  by  the  aid  of  the  table,  and  then  find  the  number 
corresponding  to  the  result. 

1.   Find  the  value  of  .0631  X  7.208  X  512.72. 

By  Art.  455,  log  (.0631  x  7.208  x  512.72)  =  log  .0631 
+  log  7.208  +  log  512.72 

log  .0631=   8.800029-10 
log  7.208=   0.857815 
log  512.72=   2.709880 


Adding,  .-.  log  of  Ans.  =  12.367724  -  10 

=    2.367724  (Art.  466,  1) 
Number  corresponding  to  2.367724  =  233.197,  Ans. 

0    „.    ,  „         .        .   3368.52 
2.    Find  the  value  of  -^^g. 

log  HJIJH  =  log  3368.52  -  log  7980.04 

log  3368.52  =  13.527439  -  10  (Art.  466,  2) 
log  7980.04=    3.902005 


Subtracting,  .-.  log  of  Ans.  =    9.625434—10 
Number  corresponding  =.422118,  Ans. 


LOGARITHMS.  355 

3.   Find  the  value  of  (.0980937)5. 

log  (.0980937)5  =  5  x  log  .0980937 
log  .0980937  =    8.991641  - 10 

5 


Multiplying,  .-.  log  of  Ans.  =  44.958205  -  50 

=   4.958205-10 
Number  corresponding  =  .0000090825,  Ans. 

4.  Find  the  value  of  ^2.36015. 

log  ^  2.3601 5  =  *  log  2.36015 

log  2.36015  =  0.372940 
Dividing  by  7,  .-.  log  of  Ans.  =  0.053277 

Number  corresponding  =  1.13052,  Ans. 

2  v^5 

5.  Find  the  value  of  — — . 

3* 

log      $    -  ]°g  2  +  I  log  5-5  log  3 
o 

log  2  =  0.301030 

log  5  =  0.698970  ;  divide  by  3  =  0.232990 

log  3  =  0.477121  0.534020 

Multiply  by  5,  =  2.385605 ;  divide  by  6  =  0.397601 
Subtracting,  .-.  log  of  Ans.  =  0.136419  • 

Number  corresponding  =  1.36905,  Ans. 

Note.     The  work  of  the  next  two  examples  will  be  exhibited  in  the 
customary  form,  the  —  10's  being  omitted  after  the  mantissse.    See  Art.  451. 


6.    Find  the  value  of  ^.00003591. 


356  ALGEBRA. 


log  {/  .00003591  =  ^  log  .00003591 

log  .00003591  =  5.555215 
7)5.555215 
log  of  Ans.  Z  9.365031  (Art.  466,  4) 
Ans.  =  .231756. 


m     -n-     ",     i  i  r       // -032956  \ 

7.   Find  the  value  of  W  ( - 


7.96183/' 


los  \J  (^SiH) =  \  ^  -03295G  " log  7,96183) 


log  .032956  =  8.517934 


log  7.96183  =  0.901013 


2)7.616921 
log  of  Ans.  =  8.808460 
Ans.  =  .0643369. 

Note.  In  computations  by  logarithms,  negative  quantities  are  used  as 
if  they  were  positive  ;  the  sign  of  the  result  being  determined  irrespective 
of  the  logarithmic  work. 

EXAMPLES. 
468.     Calculate;  by  logarithms,  the  values  of  the  following : 

1.   9.23841  x  .00369822.  5.  ^3. 

* 

3.70963  x  286.512  g      ,g 

1633.72  *  '   V 

3.  (23.846-t)8.  7.  ^5. 

4.  (- .0009296S7)*.  8.   ^.0042937. 


LOGARITHMS. 


357 


18 


9.   V-  6829.586. 


112 


10.    (1.05624) 


11.    (- .0020001G)i£. 


12.  2?  x  (-  3)*. 


13. 


14. 


3 

5T 


(-2)* 

3^ 
(-  4)§ 


15.  m 


ii 


16.   V  7239.812. 


17.   V  .00230508. 


19. 


35 

113 


/  .0872635  U 
\  .132088  /  " 


«.  i/\- 


22.   i> 


23 


21 
13* 


24.    f2x('3xf4. 

//  3258.826 \ 
V  V  49309.8  )  ' 


25 


/-  31.6259  W 
'    V  429.0162 


27_  (625.343)- 


(.732465) 


t 


28. 


29. 


30. 


V  .000128883 
y. 000827606* 

(_  .746892) ^ 
-  (.234521)^ 

ty  .00730007 

"    * 

(.682913)  ^ 


18.   V-  .000009506694.         31. 


y  5.95463  x  V  61.1998 


V  298.5434 
32.   (538.217  x  .000596899)^. 


33.  - 


304.698  x  .9026137 


.00776129  X- 16923.24 


34.   (18.9503)11  x  (-.213675)14. 


358  ALGEBRA. 


A  Orro  I    Q 


35.    V  3734.89  x  .00001108184. 


36.    (2.03172)*  x  (.712719)* 


y-  .00819323  x  (.0628513) * 
-  .9834171  " 


6/^T7T7777^T-  .8/ 


38.    \/.035  x  V  .626671  X  V-M721033. 

EXPONENTIAL  EQUATIONS. 

469.  An  Exponential  Equation  is  one  in  which  the  un- 
known quantity  occurs  as  an  exponent. 

To  solve  an  equation  of  this  form,  take  the  logarithms  of 
hoth  members  according  to  Art.  457';  the  result  will  be  an 
equation  which  can  be  solved  by  ordinary  algebraic  methods. 

1.  Given  31*  =  23  ;  find  the  value  of  x. 
Taking  the  logarithms  of  both  members, 

log  (31*)  =  log  23 
or,  by  Art.  457,  x  log  31  =  log  23 

mi  loS23      1-361728       MWVr-    , 

Whence,  X  =  ^  =  -^—^  =  .91307  < ,  Ans. 

The  value  of  the  fraction    '.^~^    may  be  obtained  bv  di- 

1.491362         J  J 

vision,  or  better  by  logarithms,  as  in  Art.  468. 

2.  Given  .2*  =  3  ;  find  the  value  of  x. 
Taking  the  logarithms  of  both  members, 

x  log  .2  =  log  3 

log  3  .477121  .477121 

\Y  hence,         x  = 


'  log  .2      9.301030  — 10  .698970 

We  may  find  the  value  of  the  fraction  by  logarithms  exactly 
as  if  it  were  positive,  and  prefix  a  —  sign  to  the  result.     Thus. 


LOGARITHMS.  359 

log  .477121  =  9.678628  - 10. 
log  .698970  =  9.814458  - 10 


Subtracting,  =  9.834170  -  10 

Number  corresponding  =  .682606 
Therefore,  x  =  —  .682606,  Ans. 

EXAMPLES. 

Solve  the  following  equations  : 

3.  II1  =  3.        5.   13*  =  .281.  7.   5*~8  =  82*+1. 

4.  .3r  =  .S.        6.   .703*  =  1.09604.        8.  233*  +  5  =  312*-3. 

APPLICATION   OF   LOGARITHMS  TO   PROBLEMS  IN 
COMPOUND   INTEREST. 

470.     Let  P  =  the  principal,  expressed  in  dollars. 

Let  t  =  the  interval  of  time  during  which  simple  interest 
is  calculated,  expressed  in  years  and  fractions  of  a  }*ear.  For 
instance,  if   the    interest    is    compounded  annually,  t  =  1  ;    if 

semi-annually,  t  =  -  ;  etc. 

Let  P  =  the  interest  of  one  dollar  for  the  time  t. 

Let  n  =  the  number  of  years. 

Let  Ai,  A2,  A3, be  the  amounts  at  the  ends  of  the  1st, 

2d,  3d, intervals. 

Let  A  be  the  amount  at  the  end  of  n  years. 

Then  Al  =  P  +  PP  =  P(l  +  P) 
A2  =  A1  +  A1B  =  A1(1  +  R) 

=p  (i  +  p)  (i  +  p)  =  p  (i  +  py 

A3  =  A,  +  A2P  =  A2(1  +  P) 

=  p(i  +  py2(i  +  p)  =  p(i  +  py 


300  ALGEBRA. 

71 

As  there  are  -  intervals,  the  amount  at  the  end  of  the  last, 
v 

according  to  the  law  observed  above, 

A  =*  P  (1  +  Sp. 

1.    Given  P,  t,  R,  and  n,  to  find  A. 

n 

As  A  =  P  (1  +  R)  *.,  we  have  by  logarithms, 
log  A  =  log  P  (1  +  Rp  =  log  P  +  log  (1  +  Rp 

71 

=  log  P  +  -  log  (1  +  R). 

h 

Example.  What  will  be  the  amount  of  $7,325.67  for  3 
years  9  months  at  7  per  cent  compound  interest,  the  interest 
being  compounded  quarterly  ? 

Here  P  =  7325.67,  t  =  j ,  R  —  .0175,  n  =  3f,  -  ==  15. 


log  P  =  3.864848 

i 


log  (1  +  R)  =  0.007534  ;  multiply  by  15  =  0.113010 
Adding,  . •.  log  of  A  =  3.977858 

Number  corresponding,  A  =  $  9502.93,  Ans. 

2.    Given  t,  R,  n,  and  A,  to  find  P. 

n  A 

As  A  =  P  (1  +  R)t ,  .-.  P  = ;  or,  by  logarithms, 

(1  +  R)~ 

/log  P  =  log  A  —  log  (1  +  Rp  =  log  A  —  "  log  (1  +  R). 

Example.    What  sum  of  money  will  amount  to  $  1 ,76« \.5B  at 
5  per  cent  compound  interest  in  3  years,  the  interest  being 

compounded  semi-annually  ? 


LOGARITHMS.  361 

1  n 

Here  t  =  % ,  B  =  .025,  »  =  3,  A  =  1763.55,  -  =  6. 

log  ^  =  3.246388 
log  (1  +  P)  =  0.010724;  multiply  by  6  =  0.064344 

Subtracting,  .-.  log  P  =  3.182044 

Number  corresponding  =  f>  1520.70,  Ans. 

3.    Given  P,  t,  P,  and  A,  to  find  n. 
In  Art.  470,  1,  we  sbowed  that 

log^  =  logP  +  -log(l  +  P) 

1/ 

.•.^log(l  +  P)=log^-logP 

£  (log  A  —  log  P) 
*   •   l~      log  (1  +  5)       • 

Example.  In  bow  many  years  will  $300.00  amount  to 
8  400.00  at  6  per  cent  compound  interest,  the  interest  being 
compounded  quarterly  ? 

Here  P  =  300,  t  =  | ,  P  =  .015,  A  =  400. 

log  400  -  log  300      2.602060  -  2.477121      .124039 

.'.71  = 


4  log  1.015      ~  4  x  .006466  .025864 

=  4.83  years,  Ans. 

4.    Given  P,  t,  n,  and  A,  to  find  P. 

n 
We  sbowed,  in  Art.  470,  3,  that  -  log(l  +  P)  =  log^-logP 

V 

1         /1     ,     T>\         log^  — logP 

•••  log  (1  +  P) ■  • 

Example.  If  $  500.00  at  compound  interest  amounts  to 
$689.26  in  6  years  and  6  months,  the  interest  being  com- 
pounded semi-annually,  what  is  tbe  rate  per  cent  per  annum  '? 


G2 


ALGEBRA. 


1  n 

Here  P  =  500,  t  =  7>,n  =  6h,A  =  689.26,  -  =  13. 

z  t 

,      ,-       ™      log  689.26- log  500 


13 


log  689.26  =  2.838383 
log  500      =  2.698970 


Subtracting, 


=  0.139413 
Dividing  by  13,  .-.  log  (1  +  B)  =  0.010724 

Number  corresponding  =  1.025  =  1  +  II,  or  R  =  .025. 

That  is,  one  dollar  gains  $  .025  semi-annually ;  or  the  rate  is 
5  per  cent  per  annum. 


EXPONENTIAL  AND  LOGARITHMIC  SERIES. 
471.     We  know  that  for  any  values  of  n  and  x, 


1  + 


x  /  1  \  n 

=  (1  +  n, 


Expanding  by  the  Binomial  Theorem,  we  obtain 


1      n  (n  -  1)   1       n  (n  -  1)  (n  -  2)    1 

1  +  n  -  H r?i 5  H rr; ?  + 


w 


w." 


[3 


?r 


.  1      v  x  (nr—1)  1       ?? x(nx— V)(nx— 2)  1 

w  2  ?r  3  ?r 


+ 


or, 


!_!      (l-l)(!-2) 

T77-  T^ +  


L3 


x  \x x  \x 

V        n  I  n 

=  1  +  x  H r^ + 


11 


\1 


; + 


LOGARITHMS.  363 

This  is  true  for  all  values  of  n ;  hence,  it  is  true  however 
large  n  may  he.     Suppose  n  to  he  indefinitely  increased.    Then 

1     2 

the  limiting  values  of  the  fractions  - ,  — ,  etc.,  are  0  (Art.  210). 

n     n 

Hence,  at  the  limit,  we  have, 
1        1 


1  +  1  +  777  +  T7V  + 


-f^  +  Sr  + 


[3_  J  '        '    [2    '   [3 

The  series  in  the  bracket  we  denote  by  e ;  hence, 


2  3 

x        x 


<?  =  !  +  *  +  -  +  -  + 


472.     To  expand  ax  in  poivers  of  x. 

Let  a  =  em  ;  whence  (Art.  444),  7ft  =  logc  ft. 


m2  a?2      ra3  x3 


Then  a*  =  era  x  =  (Art.  471)  1  +  m  a;  +  -y^-  +  -r^-  + 


Substituting  the  value  of  m, 


V  2  /I  \  *\ 


a*  =  1  +   (l0ge  ft)  X  +  (log,  ft)2  —  +   (loge  ft)3  j^   +   

This  result  is  called  the  Exponential  Theorem. 

473.  The  system  of  logarithms  which  has  e  for  its  base, 
is  called  the  Napierian  System,  from  Napier,  the  inventor  of 
logarithms.  The  value  of  e  may  be  easily  calculated  from  the 
series  of  Art.  471,  and  will  be  found  to  be  2.7182818 

474.  To  expand  log,,  (1  +  x)  in  poivers  ofx. 

a*  =  {l+(a-l)}*  =  l  +  x(al-l).+  X(x~1)(a-iy 

+  g(s-l)(s-2)  + 


=  l  +  x  {(ft-  1)  -  -(-^  +  ^^ }  +  terms  con- 

taining  x2,  xs}  etc. 


364  ALGEBRA. 

But  (Art.  472),  ax  =  1  +  x  (loge  a)  +  terms  containing  x% 

As  the  two  values  of  ax  are  equal  for  all  values  of  x,  by  the 
Theorem  of  Undetermined  Coefficients  the  coefficients  of  x  in 
the  two  expressions  are  equal;  hence, 

logea=(a  —  l) ^ 1 3 

Putting  a  =  1  +  x,  and  therefore  a  —  1  =  x,  we  obtain 

Syt"  /ytO 

loge(l  +  «)=»  —  — +  — — 


Note.  This  formula  might  be  used  to  calculate  Napierian  logarithms  ; 
but  unless  x  is  a  very  small  fraction,  the  series  in  the  second  number  is 
either  divergent  or  converges  very  slowly,  and  hence  is  useless  in  most 
cases. 

475.  To  obtain  a  more  convenient  formula  for  calculating 
the  Napierian  logarithm  of  a  number. 

X2         Xs         X*         X5 


By  Art.  4,4,     loge  (1  +  x)  =  x  —  y  +  y  —  y  +  y  — 

put  X  =  —  X, 


X2       X3        X*        Xs 


.-.loge(l  —  x)=—x 


Subtracting, 


2  x3     2xh 
• .  loge  (1  +  x)  -  log,  (l-x)  =  2x+  —  +-ir  + 


or,  by  Art.  456,  log,,  f  j 


1  +  x\      c  (         x3      x5 

>  =  2  [x+-7r+-Tr  + 


Let  x  = 


x)  3       5 

1 


1  + 


2n  +  l 
1 


l+x_  2n  +  l_2n +  1  +  1  _2n  +  2  __n  +  l 

l—x~~7       ~T~      ~2n  +  l  —  l~      2  u  n 

2n  +  l 


LOGARITHMS.  365 

Substituting,  .\  log,  p— J  =  loge  (n  + 1)  —  loge » 

~~  \2n  +  1^3(2n  +  iy  ^  5  (2  n  +  l)3  T  "      / 

■■■^ge(n+l)=\ogen+2(^  +  3^+1),+  5{:Jn+1)b+ ) 

476.     To  calculate  loge  2,  put  n  =  1  in  the  formula  of  Art. 
475. 

/     1                1                    1  \ 

...loge2  =  logeH-2^2TI+3(2  +  1)s+5^2q:-i)5+ j 

or,  since  loge  1  =  0, 

/l      1  1  1  1  1 

log. 2  =  2^3 +  81 +  12i5 +  15309  + 177147 +  1948617"1""" 


=  2  (.3333333  +  .0123457  +  .0008230  +  .0000653 
+  .0000056  +  .0000005  + ) 

=  2  x  .3405734  =  .6931468  =  .693147,    correct    to    the 
sixth  decimal  place. 

From  log,,  2,  we  may  calculate  loge  3  ;  and  so  on.  We  shall 
find  loge  10  =  2.302585. 

477.  Tn  calculate  the  common  logarithm  of  a  number  from 
its  Napierian  logarithm. 

By  Art.  460,  changing  b  to  10,  and  a  to  e,  we  obtain 

"»■ "  =  feS  =  &3S3«i  b~  »  =  -434-0945  X  ,0g'  * 
For  instance,  logI0  2  =  .4342945  x  .693147  =  .301030. 

The  multiplier  by  which  logarithms  of  any  system  are  de- 
rived from  the  Napierian  system,  is  called  the  modulus  of  that 
system.  Hence,  .4342945  is  the  modulus  of  the  common  sys- 
tem. 


366  ALGEBRA. 

As  tables  of  common  logarithms  are  met  with  more  fre- 
quently than  tables  of  Napierian,  a  rule  for  changing  common 
logarithms  into  Napierian  may  be  found  convenient. 

RULE. 

Divide  the  common  logarithm  by  .4342945. 

For  example,  to  find  the  Napierian  logarithm  of  586.324, 

common  log  586.324  =  2.768138 

Divide  by  .4342945,  .-.  Napierian  log  586.324  =  6.373873,  Ans. 

Another  method  would  be  to  multiply  the  common  logarithm 
by  2.302585,  the  reciprocal  of  .4342945. 

Napierian  logarithms  are  sometimes  called  hyperbolic  loga- 
rithms, from  having  been  originally  derived  from  the  hyper- 
bola. They  are  also  sometimes  called  natural  logarithms, 
from  being  those  which  occur  first  in  the  investigation  of  a 
method  of  calculating  logarithms.  Napierian  logarithms  are 
seldom  used  in  computation,  but  occur  frequently  in  theoretical 
investigations. 


ARITHMETICAL   COMPLEMENT. 

478.     The  Arithmetical  Complement  of  the  logarithm  of 
any  quantity  is  the  logarithm  of  the  reciprocal  of  that  quantity. 

For  example,  if  log  4098  =  3.612572,  then 

ar.  co.  l#g  4098  =  log  — —  =  log  1  -  log  4098 
°  4098         °  ° 

=  0  -  3.612572  =  6.387428  -  10. 
Again,  if  log  .06689  =  8.825361  -  10,  then 

ar.  co.  log  .06689  =  log  --?— -  =  0  -  (8.825361  -  10) 
=  10  -  8.825361  =  1.174639. 


LOGARITHMS.  367 

The  following  rules  will  be  evident  from  the  preceding 
illustrations  : 

To  find  the  arithmetical  complement  of  a  positive  loga- 
rithm, suit  nut  it  from  10,  writing  — 10  after  the  mantissa. 

To  find  the  arithmetical  complement  of  a  negative  loga- 
rithm, subtract  that  portion  of  it  besides  the  — 10  from  10. 

The  only  application  of  this  is  to  exhibit  the  work  of  calcu- 
lation by  logarithms  in  a  more  compact  form  in  certain  cases. 
It  depends  on  the' principle  that  subtracting  a  logarithm  or 
adding  its  arithmetical  complement  gives  the  same  result. 

For,  suppose  we  are  to  calculate by  logarithms. 

.      a  X  b       .       (         .       1       1 

l0«7^  =  l0glax('xcx(z 


=  log  a  +  log  b  +  log  -  +  log 


1      i      1 

c  +  ]°Sd 


=  log  a  -f-  log  b  +  ar.  co.  log  c  +  ar.  co.  log  d. 

That  is,  the  work  can  be  exhibited  in  the  form  of  the  addi- 
tion of  four  logarithms,  instead  of  the  subtraction  of  the  sum 
of  two  logarithms  from  the  sum  of  two  others.  The  principle 
is  only  applicable  to  the  case  of  fractions ;  and  the  rule  to  be 
used  is, 

Add  together  the  logarithms  of  the  quantities  in  the  numer- 
ator, and  the  arithmetical  complements  of  the  logarithms  of 
the-  quantities  in  the  denominator. 

Example.     Calculate  the  value  of  „    't,       — _100. 
1  613.8  x  .0*  .23 

log  Qi^xljfn  =  l0S  79'23  +  lQg 10-39  +  ar"  co-  IoS  613-8 

+  ar.  co.  log  .07723 


368  ALGEBRA. 

log  79.23=  1.898890 

log  10.39  =  1.016616 

ar.  co.  log  G13.8  =  7.211973  - 10 

ar.  co.  log  .07723=  1.112211 

Adding,         .-.log  of  Ans.  =  11.239693 -  10  =  1.239693 
Number  corresponding  =  17.3657,  Ans. 

Note.  The  arithmetical  complement  may  be  calculated  mentally  from 
the  logarithm,  by  subtracting  the  last  significant  figure  from  10,  and  all  the 
others  from  9. 


MISCELLANEOUS  EXAMPLES. 
479.     1.   Find  log3  2187.     (See  Art.  111.) 

2.  Find  log5 15625. 

3.  Find  the  logarithm  of  -rr  to  the  base  —2. 

*  61 

4.  Find  the  logarithm  of  —  to  the  base  8. 

5.  Find  the  characteristic  of  log2  183. 

6.  Find  the  characteristic  of  log5  4203. 

7.  Given  log  2  =  .301030,  how  many  digits  are  there  in 
217? 

8.  Given  log  3  =  .477121,  how  many  digits  are  there  in 
3^? 

9.  Findlog1356.     (See  Art.  460.) 

10.  Find  log8 163. 

11.  Find  loggo  411. 

12.  What  sum  of  money  will  amount  to  §  8705.50,  in  7 
years,  at  7  per  cent  compound  interest,  the  interest  being  com- 
pounded annually  ? 


GENERAL  THEORY  OF  EQUATIONS.      360 

13.  In  how  many  yours  will  a  sum  of  money  double  itself  at 
6  per  cent  compound  interest,  the  interest  being  compounded 
semi-annually  ? 

14.  What  will  be  the  amount  of  $1000.00  for  38  years 
3  months,  at  6  per  cent  compound  interest,  the  interest  being 
compounded  quarterly  ? 

15.  At  what  rate  per  cent  per  annum  will  $2500.00  amount 
to  $  3187.29  in  3  years  and  6  months,  the  interest  being  com- 
pounded quarterly  ? 

16.  In  bow  many  years  will  8  9681.32  amount  to  $  15308.70 
at  5  per  cent  compound  interest,  the  interest  being  compounded 
annually  ? 

17.  Using  the  table  of  common  logarithms,  find  the  Na- 
pierian logarithm  of  52.9381  (Art.  477). 

18.  Find  the  Napierian  logarithm  of  1325.07. 
19  Find  the  Napierian  logarithm  of  .085623. 
20.    Find  the  Napierian  logarithm  of  .342977. 


XLII.  —  GENERAL  THEORY  OF  EQUATIONS. 

480.  The  general  form  of  a  complete  equation  of  the  nth 
degree  is 

xn  -\- 1~>  xn~l  +  q  xn~'2  + +  t  x2  +  u  x  +  v  =  0 

"Where  n  is  a  positive  integer,  and  the  number  of  terms  is  n  +  1. 
The  quantities  j),  q, t,  u,  v  are  either  positive  or  nega- 
tive, integral  or  fractional ;  and  the  coefficient  of  x11  is  unity. 

481.  In  reducing  an  equation  to  the  general  form,  all  the 
terms  must  be  transposed  to  the  first  member,  and  arranged 
according  to  the  powers  of  x.  If  as"  has  a  coefficient,  it  may 
be  removed  by  dividing  the  equation  by  that  coefficient. 


370  ALGEBRA. 

482.  A  Root  of  an  equation  is  any  real  or  imaginary  ex- 
pression, which,  being  substituted  for  its  unknown  quantity, 
satisfies  the  equation,  or  makes  the  first  member  equal  to  0 
(Art.  166). 

We  assume  that  every  equation  has  at  least  one  root. 

483.  An  equation  of  the  third  degree  containing  only  one 
unknown  quantity,  or  one  in  which  the  cube  is  the  highest 
power  of  the  unknown  quantity,  is  usually  called  a  cubic  equa- 
tion. 

484.  An  equation  of  the  fourth  degree  containing  only  one 
unknown  quantity  is  usually  called  a  biquadratic  equation. 


DIVISIBILITY   OF   EQUATIONS. 

485.  If  a  is  a  root  of  an  equation  in  the  form 

xn  +qi  xn~l  +  q  xn~2  + +  tx2  +  ux  +  V  —  0, 

then  the  first  member  is  divisible  by  x  —  a. 

It  is  evident  that  the  division  of  the  first  member  by  x  —  a 
maybe  carried  on  until  x  disappears  from  the  remainder.  Let 
Q  represent  the  quotient,  and  R  the  remainder,  which  is  inde- 
pendent of  x ;  then  the  given  equation  may  be  made  to  take 

the  form 

.      (x  -  a)  Q  +  B  =  0. 

But  if  x  =  a,  then  (x  —  a)  Q  =  0,  and,  consequently, 

R  =  0; 

that  is,  x  —  a  is  a  factor  of  the  first  member  of  the  given  equa- 
tion, as  it  is  contained  in  it  without  a  remainder. 

486.  Conversely,  if  the  first  member  of  am.  equation  in  the 
form 

* 

xn  -\-pxn~1-\-  qxn~2  +  +  t  x2  +  ux  +  v  =  0 

is  divisible  by  x —  a,  then  a  is  a  root  of  the  equation. 


GENERAL  THEORY  OF  EQUATIONS.      371 

For,  if  the  first  member  of  the  given  equation  is  divisible  by 
x  —  a,  then  tbe  equation  may  be  made  to  take  the  form 

(x  —  a)  Q  =  0  ; 

and  it  follows  from  Art.  330  that  a  is  a  root  of  this  equation. 

EXAMPLES. 

By  the  method  of  Art.  486, 

1.  Prove  that  3  is  a  root  of  the  equation 

x3  —  6  x2  +  11  x  —  6  =  0. 

2.  Prove  that  —  1  is  a  root  of  the  equation  x3  +  1  =  0. 

3.  Prove  that  1  is  a  root  of  the  equation 

x3  +  x2  — 17  a; +  15  =  0. 

4.  Prove  that  —  2  is  a  root  of  the  equation 

a;4  —  3  x2  +  4  x  +  4  =  0. 

5.  Prove  that  4  is  not  a  root  of  the  equation 

x*  —  5  x3  +  5  x2  +  5  x  -  6  =  0. 

NUMBER  OF  ROOTS. 

487.     Every  equation  of  the  nth  degree,  containing  but  one 
unknown  quantity,  has  n  roots,  and  no  more. 

Let  a  be  a  root  of  the  equation 

xn  +2)  xn~x  +  qxn~2+ +  tx2  +  ux  +  v  —  0; 

then,  by  Art.  485,  the  first  member  is  divisible  by  x  —  a,  and 
the  equation  may  be  made  to  take  the  form 

(x  —  a)  (V1-1  +  2h  xn~2  +  +  «j  x  +  i\)  =  0. 

The  equation  may  be  satisfied  by  making  either  factor  of 
the  first  member  equal  to  0  (Art.  330)  ;  hence, 

x •  —  a  =  0 

and  x»~1+p1xn-i+ +ulx  +  v1  =  0.  (1) 


372  ALGEBRA. 

But  equation  (1)  must  have  some  root,  as  b,  and  may  be 
placed  under  the  form 

(x  —  b)  (x11-2  +jh  xn~3  + +  v,x  +  v,)=0', 

which  is  satisfied  by  placing  either  factor  of  the  first  member 
equal  to  0 ;  and  so  on. 

Since  each  of  the  factors  x  —  a,  x  —  b,  etc.,  contains  only 
the  first  power  of  x,  it  is  evident  that  the  original  equation  can 
be  separated  into  as  many  such  binomial  factors  as  there  are 
units  in  the  exponent  of  the  highest  power  of  the  unknown 
quantity,  and  no  more  ;  that  is,  into  n  factors,  or 

(x  —  a)  (x  —  b)  (x  —  c) (x  —  T)  =  0. 

Hence,  by  Art.  330,  the  equation  has  the  n  roots  a,  b,  c, /. 

Moreover,  if  the  equation  had  another  root,  as  r,  then  it 
must  contain  another  factor  x  —  r,  which  is  impossible. 

488.  It  should  be  observed  that  the  n  binomial  factors  of 
which  the  general  equation  of  the  nth  degree  is  composed,  are 
not  necessarily  unequal  j  hence,  two  or  more  of  the  roots  of  an 
equation  may  be  equal.     Thus,  the  equation 

x*  _  g  x2  +  12  x  _  8  =  0 

may  be  factored  so  as  to  take  the  form 

(x-2)(x-2)  (.r-2)=0,  or  (*-2)3  =  0; 
and  hence  the  three  roots  are  2,  2,  and  2. 

489.  It  will    be  readily   seen   that   any  equation,  one   of 

whose  mots  is  known,  may  be  depressed  to  another  of  the  next 
lower  degree,  which  shall  contain  the  remaining  roots.  Hence, 
if  all  the  roots  of  an  equation  are  known  excepl  two,  those 
may  be  obtained  from  the  depressed  equation,  hy  the  rules  for 
quadratics. 

1.  One  root  of  the  equation  xs  +  2  x1  —  23  x  —  60  =  0  is  —  3 ; 
what  are  the  others  ? 


GENERAL  THEORY  OF  EQUATIONS.      373 

Dividing  x3  +  2  x~  —  23  x  —  60  by  x  +  3,  the  given  equation 
may  be  put  in  the  form 

O  +  3)O2-x-20)=0. 

Thus,- the  depressed  equation  is  x2  —  x  —  20  =  0. 

Solving  this  by  the  rules  for  quadratics,  we  obtain  x  =  5  or 

—  4  ;  which  are  the  remaining  roots. 

EXAMPLES. 

2.  One  root  of  the  equation  xz  —  19  x  +  30  =  0  is  2  ;  what 
are  the  others  ? 

3.  Eequired   the    three    roots  of    the    equation  Xs  =  a3,  or 

x3  —  a3  —  0. 

4.  One  root  of  the  equation  x3  +  x2  —  16  x  +  20  =  0  is  —  5  ; 
reqxiired  the  remaining  roots. 

5.  Two  roots  of  the  equation  xA  —  3  x3  — 14  x"  +  48  x  —  32  =  0 
are  1  and  2  ;  required  the  remaining  roots. 

6.  One  root   of  the  equation  x4  —  7  x3  +  3  x  +  3  =  0  is  1  ; 
what  equation  contains  the  remaining  roots  ? 

7.  One  root  of  the  equation  6  x3  —  x~  —  32  x  +  20  =  0  is  2 ; 
what  are  the  others  ? 

8.  Two  roots  of  the  equation  20  x*  -  169  x3  +  192  x~  +  97  x 

—  140  =  0  are  1  and  7  :  what  are  the  others  ? 


5 


FORMATION    OF  EQUATIONS. 

490.  An  equation  having  any  given  roots  may  be  formed 
by  subtracting  each  root  from  the  unknown  quantity,  and  pla- 
cing the  product  of  these  binomial  factors  equal  to  0. 

For  it  is  evident,  from  principles  already  established,  that 

an  equation  having  the  n  roots  a,  b,  c,  I  may  be  written 

in  the  form 

(x  —  a)  (x  —  b)  (x  —  c)  (x  —  I)  =  0. 


374  ALGEBRA. 

After  performing  the  multiplication  indicated,  the  equation 
will  assume  the  form  . 

xn  +2?xn~1  +  qxn~2  + +  tx2  +  ux  +  v  =  0. 

(Compare  Art.  329.) 

1.  Form  the  equation  whose  roots  are  1,  2,  and  —  4. 
Result,  (x  —  1)  (cc  —  2)  (cc  +  4)  =  0 

or,  x3  +  x2  — 10  x  +  8  =  0. 

EXAMPLES. 

Form  the  equations  whose  roots  are : 

2.  -  1,  -  3,  and  -  5.  6.    1,  2,  3,  and  4. 

3.  5,  -  2,  and  -  3.  7.   4,  4,  and  5. 

4.  1,  - ,  and  - .  8.   0,  -  1,  3,  and  4. 

5.  ±  1  and  ±2.  9.   -  5  ? ,  -  2,  and  ? 

4  3 


COMPOSITION  OF  COEFFICIENTS. 

491.  The  coefficient  of  the  second  term  of  an  equation  of  the 
nth  degree  in  its  general  form  is  the  sum  of  all  the  roots  with 
their  signs  changed;  that  of  the  third  term  is  the  sum  of  their 
products,  taken  two  and  two  ;  that  of  the  fourth  term  is  the 
sum  of  their  'products,  taken  three  and  three  with  their  .signs 
changed,  etc. ;  and  the  last  term  is  the  product  of  all  the  roots 
with  their  signs  changed. 

For,  resuming  the  equation 

(x '—  a)  (x  —  b)(x  —  c) (x  —  k)  (x  —  l)=0, 

if  we  perform  the  multiplication  indicated,  we  obtain 

(x  —  a)  (x  —  b)  =  x~  —  (a  +  b)  x  +  a  b, 
(x—a)(x  —  b)  (x—c)  =  x3—(a  +  b+c)x+(ab  +  ac+bc)x—abc, 


GENERAL  THEORY  OF  EQUATIONS.     375 

and  so  on.     When  n  factors  have  heen  multiplied,  the  coeffi- 
cients of  the  general  equation  become 

jp  =  —  a  —  b  —c— —  k  —  I 

q  =  ab  +  ac-\-bc-\- +  kl 

r  =  —  a  b  c  —  ab  d  —  a  c  d  — —  ikl 


v  =  ±a  b  c kl 

which  corresponds  with  the  enunciation  of  the  proposition ; 
the  upper  sign  of  the  value  of  v  being  taken  when  n  is  even, 
and  the  lower  sign  when  n  is  odd. 

492.  If  ^=0,  that  is,  if  the  second  term  of  an  equation 
be  wanting,  the  sum  of  the  roots  will  be  0. 

If  v  =  0,  that  is,  if  the  absolute  term  of  an  equation  be  want- 
ing, at  least  one  root  must  be  0. 

493.  Every  rational  root  of  an  equation  is  a  divisor  of  the 
last  term. 

494.  When  all  the  roots  of  an  equation  but  two  are  known, 
the  coefficient  of  the  second  term  of  the  depressed  equation 
(Art.  489)  can  be  found  by  subtracting  the  sum  of  the  known 
roots,  with  their  signs  changed,  from  the  coefficient  of  the 
second  term  of  the  original  equation.  The  absolute  term  of 
the  depressed  equation  can  be  found  by  dividing  the  absolute 
term  of  the  original  equation  by  the  product  of  the  known 
roots  with  their  signs  changed. 

EXAMPLES. 

Find  the  sum  and  product  of  the  roots  in  the  following : 

1.   a;3-7a;+6  =  0.       2.   2  x*-  5  xs  -  17  x2  +  14 x  +  24  =  0. 

In  the  following  example  obtain  the  depressed  equation  by 
the  method  of  Art.  494  : 

3.  Two  roots  of  the  equation  x4  —  5  xs  —  2  x2  +  12  x  +  8  =  0 
are  2  and  —  1 ;  what  are  the  others  ? 


376  ALGEBRA. 


FRACTIONAL   ROOTS. 

495.  An  equation  whose  coefficients  are  all  integral,  the 
coefficient  of  the  first  term  being  unity,  cannot  have  a 
rational  fraction  as  a  root. 

If  possible,  let  -,  a  rational  fraction  in  its  lowest  terms,  be 
a  root  of  the  equation 

xn  +  pxn~l  +  qxn~2  + +  tx2  +  UX  +  v  =  0, 

where  p,  q, ,  t,  u,  v  are  integral.     Then 

a\n         faV'1  fa\n-2  ( a\2         fa 


Multiplying  through  by  lf~l,  and  transposing, 

-n  =  —  (2,an-1+qan-2b  + +  ta2bn~s  +  ua  un~2+  vbn~l). 

Now,  as  -  is  in  its  lowest  terms,  a  and  b  can  have  no  com- 
mon divisor ;  therefore  an  and  b  can  have  no  common  divisor ; 

a" 
hence  —  is  in  its  lowest  terms.     Thus,  we  have  a  fraction  in 
o 

its  lowest  terms  equal  to  an  entire  quantity,  which  is  impossi- 
ble. Therefore  no  root  of  the  equation  can  be  a  rational 
fraction. 

Note.     The  equation  may  have  an  irrational  fraction  as  a  root,  such  as 

'     for  example.     Such   a  root,   whose  value  can  only   be  expressed 

4 
approximately  by  a  decimal  fraction,  is  called  incommensurable. 


IMAGINARY   ROOTS. 

496.     If  the  coefficients  of  an  equation  be  real  quantities, 
imaginary  roots  enter  it  by  pairs,  if  at  all. 


Suppose  a  +  b  ^  —  1  to  be  a  root  of  the  equation 

Xn  +  p  X"  ~  l  +  q  Xn  ~  2  + +  t  X2  +  U  X  +  V  =  0. 


GENERAL  THEORY  OF  EQUATIONS.      377 


Substituting  a  +  b  y/—  1  for  x,  and  developing  each  expres- 
sion by  the  Binomial  Theorem,  all  the  odd  terms  of  each  series 
will  contain  either  powers  of  a,  or  even  powers  of  b  y/ —  1,  and 
are  therefore  real ;  while  all  the  even  terms  contain  the  odd 
powers  of  b  y — 1,  and  are  therefore  imaginary.  Representing 
the  sum  of  all  the  real  quantities  by  P,  and  the  sum  of  all 
the  imaginary  quantities  by  Q  y —  1,  we  have 


P+  £y/-l=0. 

This  equation  can  be  true  only  when  both  P  and  Q  equal  0. 

If  we  now  substitute  a  —  b  y/ —  1  for  x,  we  find  that  the 
series  differ  from  the  former  only  in  having  their  even  or 
imaginary  terms  negative.  Hence,  we  obtain  as  the  first 
member 


P-Q^-l, 

which  must  be  equal  to  0,  for  we  have  already  shown  that  both 
P  and  Q  equal  0.     Thus,  a  —  b  y/—  1  satisfies  the  equation. 

Similarly,  we  may  show  that, if  b  y/ —  1  is  a  root  of  the  equa- 
tion, then  will  —  b  y/—  1  also  be  a  root  of  the  equation. 

497.     The    product    of  a  pair  of  imaginary   quantities   is 
always  positive.     Thus, 


(a  +  6y/-l)  (a-b^-l)  =  a2  +  b 
and  0  y/^l)  (-  b  y/^I)  =  b\ 


TRANSFORMATION   OF   EQUATIONS. 

498.    To  transform  an  equation  into  another  which  shall 
have  the  same  roots  with  contrary  signs. 

Let  the  given  equation  be 

xn  +  p  xn  ~ 1  +  q  xn  ~ 2  + +  t  x2  +  u  x  +  v  =  0. 


378  ALGEBRA. 

Put  x  =  —  y;  then  whatever  value  x  may  have,  y  will  have 
the  same  value  with  its  sign  changed.  The  equation  now 
becomes 

{-y)n+p(-y)n-l  +  q(-y)n-2+ +  t(-y)2  +  u{-y) 

+  v  =  o. 

If  11  is  even,  the  first  term  is  positive,  second  term  nega- 
tive, and  so  on ;  and  the  equation  may  be  written 

yn_pyU-\    +    (1yn~1_ +     f    y'2   _    ^    y    _J_    y    _    Q  Qj 

If  n  is  odd,  the  first  term  is  negative,  second  term  positive, 
and  so  on ;  hence,  changing  all  signs,  we  write  the  equation 

y"  —v  yn  ~ 1  + 1  yn  ~ 2  — —  ^  y1  +  uy~  v  —  o.       (2) 

From  (1)  and  (2)  it  is  evident  that  to  effect  the  desired 
transformation  we  have  simply  to  change  the  signs  of  the 
alternate  terms,  beginning  with  the  second. 

Note.  The  preceding  rule  assumes  that  the  given  equation  is  complete 
(Art.  300) ;  if  it  be  incomplete,  any  missing  term  must  be  put  in  with 
zero  as  a  coefficient. 

1.  Transform  the  equation  x3  —  7x  +  6  =  0  into  another 
which  shall  have  the  same  roots  with  contrary  signs. 

We  may  write  the  equation  x3  +  0  .  ar  —  7  x  +  6  =  0. 

Applying  the  rule, 

xs  _  o  .  x2  —  7  x  —  6  =  0,  or  xz  —  7  x  —  6  =  0,  Ans. 

EXAMPLES. 

Transform  the  following  equations  into  others  which  shall 
have  the  same  roots  with  contrary  signs: 

2.  x* - 2  xs  +  x - 1.32  =  0.  3.   x5-3x2+  8  =  0. 

499.  To  transform  an  equation  into  another  whose  roots 
shall  be  some  multiple  of  those  of  the  first. 

Let  the  given  equation  he 

xn+pxn-1  +  qxn~2  + +  tx'2  +  ux  +  v  =  0. 


GENERAL  THEORY  OF  EQUATIONS.      379 

v 
Put  x  =  — ;  then  whatever  value  x  may  have,  y  will  have 
m 

a  value  m  times  as  great.     The  equation  now  becomes 

(i)r+jjLY-\Ji\"-\ +,(£)*+.(»)+.=* 

Multiplying  through  by  ran,  we  have 
yn+pmyn~1+qm2yn~2  + +  tmn~2y2  +  umn~1y  +  vm"  =  0. 

Hence,  to  effect  the  desired  transformation,  multiply  the 
second  term  by  the  given  factor,  the  third  term  by  its  square, 
and  so  on. 

Similarl}'-,  we  may  transform  an  equation  into  one  whose 
roots  shall  be  those  of  the  first  divided  by  some  quantity. 

1.  Transform  the  equation  x3  —  7  x  —  6  =  0  into  another 
whose  roots  shall  be  4  times  as  great. 

The  equation  may  be  written,  xs  -f-  0  .  x2  —  7  x  —  6  =  0. 

Then,  by  the  rule, 

x3  -  42 .  7  x  -  43 .  6  =  0,  or  x3  -  112  x  -  384  =  0,  Ans. 

EXAMPLES. 

2.  Transform  the  equation  xz  —  2  x2  +  5  =  0  into  another 
whose  roots  shall  be  5  times  as  great. 

3  xs 

3.  Transform  the  equation  xi  -| — 27  =  0  into  another 

whose  roots  shall  be  one  third  as  great. 

500.  Ta  transform  an  equation  containing  fractional 
coefficients  into  another  whose  coefficients  are  integral,  that  of 
the  first  term  being  unity. 

If  in  Art.  499  we  assume  m  equal  to  the  least  common 
multiple  of  the  denominators,  it  will  always  remove  them ; 
but  often  a  smaller  number  can  be  found  which  will  produce 
the  same  result. 


380  ALGEBRA. 

1.    Transform    the    equation    xs  —  — —  —  +  — —  =  0    into 

o       So      lUo 

another  whose  coefficients  shall  he  integral. 

The  least  common  multiple  of  the  denominators  is  108 ;  so 
that  one  solution  would  he,  by  Art.  499, 

rf_M8.  |-.M8-:|  + 108^  =  0. 

An  easier  way,  however,  is  as  follows ;  the  denominators 
may  be  written  3,  32  X  2'2,  and  33  X  22,  so  that  the  multiplier 
3  x  2  or  6  will  remove  them.     Hence,  by  Art.  499,  we  have 

xa_6    *a__63    *  +6s     *   =0     or    x*-2x*-x  +  2  =  0, 

0  ob  106 

whose   roots   are   6   times   as    great   as   those   of   the    given 
equation. 

EXAMPLES. 

Transform  the  following  equations  into  others  whose  coef- 
ficients shall  be  integral : 

**'+T~74=0-       *.-,+  re-g-»=a 

av-^  +  fwb.       5.^-5^-^  +  5  =  0. 

no  4  I 

501.  To  transform  an  equation  into  another  ivhose  roots 
shall  be  the  reciprocals  of  those  of  the  first. 

Let  the  given  equation  be 

xn  +  p  xn  ~ 1  +  q  xn  ~  -  + +  t  x2  +  u  x  +  v  =  0. 

Put  x  =  -  ;  then  whatever  value  x  may  have,  y  will  be  its 

v 

reciprocal.     The  equation  now  becomes 

1  p  q  t        u  n 

y      y  y  y     y 


GENERAL  THEORY  OF  EQUATIONS.      381 

Multiplying  through  by  y",  and  reversing  the  order, 

vyn  +  uyn~1  +  ty"-2  + +  qy2  +py  +  1  =  0. 

Dividing  through  by  v, 

u         ,        t  a    „       7}  1 

n+  _y«-i+  -  gf-»  + +  Lf  +  Ll/+  ±=0. 

v  «  w  «;  «; 

Hence,  to  effect  the  transformation,  write  the  coefficients  in 
reverse  order,  and  then  divide  by  the  coefficient  of  the  first 
term. 

EXAMPLES. 

Transform  the  following  equations  into  others  whose  roots 
shall  he  the  reciprocals  of  those  of  the  first : 

1.  a;3_6x2+lliC-6  =  0.       3.   x3-9x2+  —  -i  =  0. 

7        49 

2.  xi~x3-3x2  +  x  +  2  =  0.    4.   x3-4x2  +  9  =  0. 

502.  To  transform  an  equation  into  another  whose  roots 
shall  differ  from  those  of  the  first  by  a  given  quantity. 

Let  the  given  equation  be 

s"+ju""'  +  qxn~2+ +  tx*  +  ux  +  v  =  0.        (1) 

Put  x  =  y  +  r,  and  we  have 

(ff  +  r)*+p(t,.  +  r)—1+ +  u  (y  +  r)  +  v=0.        (2) 

Developing  (y  +  r)n,  (y  +  r)"-\ ,  by  the  Binomial  The- 
orem, and  collecting  terms  containing  like  powers  of  y,  we 
have  an  equation  of  the  form 

yn  +pi  yn-l+  ^  y«-1  + +  ^  y2  +  ^  y  +  ^  =  Q  (o) 

As  y=x  —  r,  the  roots  of  (3)  are  evidently  less  by  r  than 
those  of  (1).  By  putting  x  =  y  —  r,  we  shall  obtain  in  the 
same  way  an  equation  whose  roots  are  greater  by  r  than 
those  of  (1). 

503.  If  n  is  small,  the  operation  indicated  in  Art.  502 
may  be  effected  with  little  trouble;  but  for  equations  of  a 
higher  degree  a  less  tedious  method  is  better. 


382  ALGEBRA. 

If  in  (3)  we  put  y  =  x  —  r,  we  shall  have 

(x—rf  +2h(x  —  r)n-1+ +u1(x  —  r)  +  v1  =  0,         (4) 

which  is,  of  course,  identical  with  (1),  and  must  reduce  to  (1) 
when  developed.     If  we  divide  (4)  by  x  —  r,  we  obtain 

(x  -  r)" ~ J  +jh  0  -  r)n -2  +  qi{x-r)"-"+ +u,      (5) 

as  a  quotient,  with  a  remainder  of  vv  Dividing  (5)  by  x  —  r, 
we  obtain  a  remainder  of  ux ;  and  so  on,  until  we  obtain  all 
the  coefficients  of  (3)  as  remainders. 

Hence,  to  effect  the  desired  transformation, 

Divide  the  given  equation  by  x  —  r  or  x  +  r>  according  as 
the  roots  of  the  transformed  equation  are  to  be  less  or  greater 
than,  those  of  the  first  by  r,  and  the  remainder  will  be  the 
absolute  term  of  the  transformed  equation.  Divide  the  quo- 
tient just  found  by  the  same  divisor,  and  the  remainder  will 
be  the  coefficient  of  the  last  term  but  one  of  the  transformed 
equation  ;  and  so  on. 

504.  1.    Transform   the   equation  x3  +  3  x2  —  4^  +  1  =  0 

into  one  whose  roots  shall  be  greater  by  1. 

Using  the  method  of  Art.  502,  put  x  =  y  —  1. 
Then,         (y-iy+3(y~iy-4(y-l)  +  l  =  0, 
or,    t/3-  3  y2  +  3  7/  -  1  +  3  /  -  6  y  +  3  -  4  y  +  4  +  1  =  0, 
or,  if  —  7  y  +  7  =  0,  Ans. 

EXAMPLES. 

2.  Transform  the  equation  x3  —  x  —  6  =  0  into  one  whose 
roots  shall  be  less  by  8. 

3.  Transform  the  equation  x*  -f  6  x3  —  x1  —  5  x  —  1  =  0 
into  one  whose  roots  shall  be  greater  by  3. 

505.  To  transform  a  complete  equation  into  one  tchos* 
second  term  shall  be  wanting. 


GENERAL  THEORY  OF  EQUATIONS.     383 

The  coefficient  of  y"-1  in  (2),  Art.  502,  is  n  r  +  p.     Hence, 
in  (3),  py  =  n  r  +  p.     To  make  ^  =  0,  it  is  only  necessary  to 

make  nr  +  p  =0,  or  r  =  —  —  ;  hence,  to   effect   the   desired 

n 

V 
transformation,   put  x  =  y  —  -  ;  that    is,  put  x  equal  to  y} 

minus  the  coefficient  of  the  second  term  of  the  given  equation 
divided  by  the  degree  of  the  equation. 

1.  Transform    the    equation    x3  —  G  x~  -j-  9  x  —  G  =  0    into 
another  whose  second  term  shall  be  wanting. 

Here  p  =  —  6,  n  =  3  ;  then,  put  x  =  y  —  — —  =  y  +  2. 

Result,       (y  +  2)3  -  6  (y  +  2)2  +  9  (y  +  2)  -  6  =  0, 

or,    tf  +  6y*  +  12y  +  8-6y*-24:y-24:+9y  +18-6  =  0, 

or,  y3  —  3  y  —  4  =  0,  Ans. 

EXAMPLES. 

Transform  the  following  equations  into  others  whose  second 
terms  shall  be  wanting : 

2.  x2-px  +  q  =  0.  4.   x3  +  6x2-3x  +  4:  =  0. 

3.  xs  +  x2  +  4  =  0.  5.   x4  -  4  x3  -  5  x  —  1  =  0. 


DESCARTES'   RULE   OF   SIGNS. 

506.  A  Permanence  of  sign  occurs  when  two  successive 
terms  of  a  series  have  the  same  sign. 

A  Variation  of  sign  occurs  when  two  successive  terms  of  a 
series  have  contrary  signs. 

DESCARTES'  RULE. 

507.  A  complete  equation  cannot  have  a  greater  number 
of  positive  roots  than  it  has  variations  of  sign,  nor  a  greater 
number  of  negative  roots  than  it  has  permanences  of  sign. 


384  ALGEBRA. 

Let  any  complete  equation  have  the  following  signs : 

+  +  +  -  +  -  + 

in  which  there  are  three  permanences  and  five  variations. 

If  we  introduce  a  new  positive  root  a,  we  multiply  this  by 
x  —  a  (Art.  490).  Writing  only  the  signs  which  occur  in  the 
operation,  we  have 


+  +  +  -  + 

— 1 

+  — 

+  +  +  -  + 

-  +  — 

+ 

-  +  —  +  + 

+  ±  ±  — +  —  +  —  ±  + 

123456789    10 

a  double  sign  being  placed  wherever  the  sign  of  a  term  is 
ambiguous. 

However  the  double  signs  are  taken,  there  must  be  at  least 
one  variation  between  1  and  4,  and  one  between  8  and  10, 
and  there  are  evidently  four  between  4  and  8 ;  or  in  all  there 
are  at  least  six  variations  in  the  result.  As  in  the  original 
equation  there  were  five  variations,  the  introduction  of  a 
positive  root  has  caused  at  least  one  additional  variation ;  and 
as  this  is  true  of  any  positive  root,  there  must  be  at  least  as 
many  variations  of  sign  as  there  are  positive  roots. 

Similarly,  hy  introducing  the  factor  x  +  a,  we  may  show 
that  there  are  at  least  as  many  permanences  of  sign  as  there 
are  negative  roots.  , 

If  the  equation  is  incomplete,  any  missing  term  must  be 
supplied  with  ±  0  as  its  coefficient  before  applying  Descartes' 
Rule. 

508.  In  any  complete  equation,  the  sum  of  the  number 
of  permanences  and  variations  is  equal  to  the  number  of  terms 
less  one,  or  is  equal  to  the  degree  of  the  equation  (Art.  480). 
Hence,  when  the  roots  are  all  real,  the  number  of  positive 
roots  is  equal  to  the  number  of  variations,  and  the  number  of 
negative  roots  is  equal  to  the  number  of  permanences  (Art.  4S7). 


GENERAL  THEORY  OF  EQUATIONS.     385 

A  complete  equation  whose  terms  are  all  positive  can  have 
no  positive  root ;  and  one  whose  terms  are  alternately  positive 
and  negative  can  have  no  negative  root. 

509.  In  an  incomplete  equation,  imaginary  roots  may 
sometimes  he  discovered  hy  means  of  the  douhle  sign  of  0  in 
the  missing  terms.     Thus,  in  the  equation 

x3  +  x2  ±  0  x  +  4  =  0 

if  we  take  the  upper  sign,  there  is  no  variation,  and  conse- 
quently no  positive  root;  if  we  take  the  lower  sign,  there  is 
hut  one  permanence,  and  hence  but  one  negative  root.  There- 
fore, as  the  equation  has  three  roots  (Art.  487),  two  of  them 
must  he  imaginary. 

In  general,  whenever  the  term  which  precedes  a  missing 
term  has  the  same  sign  as  that  which  follows,  the  equation 
must  have  imaginary  roots ;  where  it  has  the  opposite  sign, 
the  equation  may  or  may  not  have  imaginary  roots,  hut 
Descartes'  Hule  does  not  detect  them.  If  two  or  more  suc- 
cessive terms  of  an  equation  he  wanting,  there  must  be  imagi- 
nary roots. 

Note.  In  all  applications  of  Descartes'  Rule,  the  equation  must  con- 
tain a  term  independent  of  x,  that  is,  no  root  must  be  equal  to  zero  (Art. 
330)  ;  for  a  zero  root  cannot  be  considered  as  either  positive  or  negative. 

EXAMPLES. 

510.  The  roots  of  the  following  equations  being  all  real, 
determine  their  si  cms  : 


*&' 


1.  x8-3x-2  =  0.        3.   a;8 -7  a;2 +  36  =  0. 

2.  .x3-10a;  +  3  =  0.      4.   xi-2x3-13x2  +  38*-24  =  0. 

5.    What  are  the  signs  of  the  roots  of  the  equation  x3  +  x1 
-4  =  0? 

DERIVED   POLYNOMIALS. 
511.     If  we  take  the  polynomial 

a  xn  +  b  xn  ~ 1  +  c  xn  ~ 2  + 


386  ALGEBRA. 

and  multiply  each  term  by  the  exponent  of  a;  in  that  term, 
and  then  diminish  the  exponent  by  1,  the  result 

n  a  x" ~l  +  (n  —  1)  b  xH~2  +  Qi  —  2)  c  xn~3  +  

is  called  the  first  derived  polynomial  or  first  derivative  of  the 
given  polynomial. 

The  second  derived  polynomial  or  second  derivative  is  the 
first  derived  polynomial  of  the  first  derivative ;  and  so  on. 
The  given  polynomial  is  sometimes  called  the  primitwe  poly- 
nomial. 

A  derived  equation  is  one  whose  first  member  is  a  deriva- 
tive of  the  first  member  of  another. 

1.  Find  the  successive  derivatives  of  x*  +  5  x1  +  3  x  +  9. 

Result :  First,      3  x2  +  10  x  +  3. 
Second,  6  x  +  10. 
Third,     6. 
Fourth,  0. 

EXAMPLES. 

Find  the  successive  derivatives  of  the  following : 

2.  a;3-5r  +  6x-2.  4.    a  x*  -bx%  +  ex  -Sd. 

3.  2cc2-ic-7.  5.    7cc4-13a;2+8x-l. 

EQUAL   ROOTS. 
512.   Let  the  roots  of  the  equation 

xn  +pxn-'1  +  qxn~2  + +  tx2  +  ux  +  v  =  0         (1) 

be  a,  b,  c, Then  (Art.  490),  we  have 

xn  +2?  xn  ~ 1  +  q  xn  ~ 2  + =  (x  —  a)  (x  —  b)  (x  —  c) 

Putting  x  +  y  in  place  of  x, 


(x  +  y)n  +  p  (x  +  y)"-1  +  ...  =  (y  +  x  -a)  (y  +  x-b) ...       (2) 
By  Art.  399,  the  coefficient  of  y  in  the  first  member  is 

nxn~1+p  (n  —  I)  xn~2  +  q  (n  —  2)  xn~*  + (3) 


GENERAL  THEORY  OF  EQUATIONS.      387 

which,  we  observe,  is  the  first   derivative  of  (1)  ;  and,  as  in 

Art.  491,  regarding  x —  a,  x —  b, as   single    terms,   the 

coefficient  of  y  in  the  second  member  is 

(x  —  b)  (x  —  c)  (x  —  d) to  n  —  1  factors  "1 

+  (x  —  a)  (x  —  c)  (x  —  d) to  n  —  1  factors  ! 

+  (x  —  a)  (x  —  b)  (x  —  d) to  ii  —  1  factors  |  ^  ' 

+  ......  j 

As  (2)  is  identical,  by  Art.  413  these  coefficients  are  equal. 

Now  if  b  =  a,  that  is,  if  equation  (1)  has  two  roots  equal  to 
a,  every  term  of  (4)  will  be  divisible  by  x  —  a,  hence  (3)  will 
be  divisible  by  the  same  factor ;  therefore  (Art.  486)  the  first 
derived  equation  of  (1)  will  have  one  root  equal  to  a.  Sim- 
ilarly, if  c  =  b  =  a,  that  is,  if  (1)  has  three  roots  equal  to  a, 
(3)  will  have  two  roots  equal  to  a ;  and  so  on.     Or,  in  general, 

If  an  equation  has  n  roots  equal  to  a,  its  first  derived  equa- 
tion xv ill  have  n  —  1  roots  equal  to  a. 

513.  From  the  principle  demonstrated  in  Art.  512,  it  is 
evident  that  to  determine  the  existence  of  equal  roots  in  an 
equation  we  must 

Find  the  greatest  common  divisor  of  the  first  member  and 
its  first  derivative.  If  there  is  no  common  divisor  there  can 
be  no  equal  roots.  If  there  is  a  greatest  common  divisor,  by 
placing  it  equal  to  zero  and  solving  the  resulting  equation  we 
shall  obtain  the  required  roots. 

The  number  of  times  that  each  root  is  found  in  the  given 
equation  is  one  more  than  the  number  of  times  it  is  found 
in  the  equation  formed  from  the  greatest  common  divisor. 

If  the  first  member  of  the  given  equation  be  divided  by  the 
greatest  common  divisor,  the  depressed  equation  will  contain 
the  remaining  roots  of  the  original  equation. 

1.    Find  the  roots  of  the  equation 

x*  -  14  xs  +  61  x~  -  84  x  +  36  =  0. 

Here   the   first   derivative   is  4  xs  —  42  x%  +  122  x  —  84  ;  the 
greatest  common  divisor  of  this  and  the  given  first  member 


388  ALGEBRA. 

is  x1  —  7  x  +  6.  Placing  x2  —  7  x  +  6  ==  0,  we  have,  by  the 
rules  of  quadratics,  or  by  factoring,  x  =  1  or  6.  Therefore 
the  roots  of  the  given  equation  are  1,  1,  6,  and  6. 

EXAMPLES. 

Find  all  the  roots  of  the  following : 

2.  xs-Sx-+13x-6  =  0.      4.   .T4-6a;2-8a;-3  =  0. 

3.  x3-7x2+16x-12  =  Q.      5.   cc4-24x2  +  64z-48  =  0. 

514.  When  the  equation  formed  from  the  greatest  com- 
mon divisor  is  of  too  high  a  degree  to  be  conveniently  solved, 
we  may  in  certain  cases  compare  it  with  its  own  derived 
equation,  and  thus  obtain  a  common  divisor  of  a  lower  degree. 
Of  course  this  can  only  be  done  when  the  equation  formed 
from  the  greatest  common  divisor  has  equal  roots. 

For  example,  required  all  the  roots  of 

xb  -  13  x*  +  67  x3  -  171  x-  +  216  x  -  108  =  0.         (1) 

Here  the  first  derivative  is  5  x*  —  52  x3  +  201  ,x2  —  342  x 
+  216 ;  the  greatest  common  divisor  of  this  and  the  given 
first  member  is  x3  —  8  x2  +  21  x  —  18.  We  have  then  to  solve 
the  equation 

x3-8x2+21z-18  =  0.  (2) 

The  first  derivative  of  (2)  is  3  x2  —  16  x  +  21 ;  the  greatest 
common  divisor  of  this  and  x3  —  8  x2  +  21  x  —  18  is  x  —  3. 
Solving  x  —  3  =  0,  we  have  x  =  3;  hence  two  of  the  roots  of 
(2)  are  equal  to  3.  Dividing  the  first  member  of  (2)  by 
(x  —  3)2  or  by  x2  —  6  x  +  9,  the  depressed  equation  is 

x  —  2  =  0,  whence  x  =  2. 

Thus  the  three  roots  of  (2)  are  3,  3,  and  2.  Hence,  the  five 
roots  of  (1)  are  3,  3,  3,  2,  and  2. 

515.  If  an  equation  has  two  roots  equal  in  magnitude,  but 
opposite  in  sign,  by  changing  the  signs  of  the  alternate  terms 
beginning  with  the  second  we  shall  obtain  an  equation  with 
these  same  two  roots  (Art.  498)  ;  then  evidently  the  greatest 


GENERAL  THEORY  OF  EQUATIONS.      389 

common  divisor  of  the  two  first  members  placed  equal  to  zero 
will  determine  the  roots. 

For  example,  required  all  the  roots  of 

x*  +  3  x3  -  13  x--21  x  +  36  =  0.  (1) 

Changing  the  signs  of  the  alternate  terms,  we  have 

x4  -  3  x3  - 13  x°-  +  27  x  +  36, 

the  greatest  common  divisor  of  which  and  the  given  first 
member  is  x2  —  9  ;  solving  x2 —  9  =  0,  we  have  x  =  3  or  —  3. 
thus  giving  two  of  the  roots  of  (1).  Dividing  the  first  mem- 
ber of  (1)  by  x2  —  9,  we  have  for  the  depressed  equation 

x-  +  3  x  -  4  =  0, 

whence  x  =  1  or  —  4.  Thus  the  roots  of  (1)  are  3,  —  3,  1, 
or  —4. 

LIMITS   OF  THE   ROOTS   OF  AN   EQUATION. 
516.     A  polynomial  of  the  form 

xn  +  p  xn  ~  *  +  q  xn  ~  2  +  Jrtx2+  ux+v 

which  we  shall  represent  by  X,  may  also  be  expressed  thus 
(Art.  490)  : 

(x  —  a)  (x  —  b)  (x  —  c)  (x  —  I)  Y 

in  which  a,  b,  c, I  are  the  real,  unequal  roots  of  the  equa- 
tion X=  0,  in  the  order  of  their  magnitude,  a  being  algebrai- 
cally the  smallest;  and  Y the  product  of  all  the  factors  con- 
taining imaginary  roots,  which  must  always  be  positive,  and 
cannot  affect  the  sign  of  X,  for  each  pair  of  imaginary  roots 
(Art.  497)  produces  a  positive  factor. 

Suppose  x  to  commence  at  any  value  less  than  a,  and  to 
assume  in  succession  all  possible  values  up  to  some  quantity 
greater  than  I.  When  x  is  less  than  a,  each  of  the  factors 
x  —  a,  x  —  b, is  negative,  and  therefore  X  is  either  posi- 
tive or  negative,  according  as  the  degree  is  even  or  odd. 


390  ALGEBRA. 

When  x  ==  a,  X  =  0.  When  x  is  greater  than  a,  and  less 
than  b,  x  —  a  becomes  positive,  and  the  sign  of  X  changes. 
Also,  when  the  value  of  x  is  made  equal  to  b,  and  then  greater, 
X first  becomes  0  and  then  changes  sign;  and  so  on,  for  each 
real  root. 

When  x  has  any  value  greater,  than  I,  X  must  be  positive ; 
for  all  its  factors  are  positive. 

517.  If  two  numbers,  when  substituted  for  the  unknown 
quantity  in  an  equation,  give  results  having  a  different  sign, 
at  least  one  root  lies  between  those  numbers. 

It  is  evident,  from  Art.  516,  that  if  X  has  a  different  sign 
for  two  values  of  x,  some  odd  number  of  roots  lies  between 
them. 

When  the  numbers  substituted  differ  by  unity,  it  is  evident 
that  the  integral  part  of  the  root  is  known. 

EXAMPLES. 

1.  What  is  the  first  figure  of  a  root  of  the  equation  x3  +  3  x- 

Here,  if  x  =  2,  the  first  member  becomes  —  2  ;  and  if  x  =  3, 
the  first  member  becomes  25 ;  therefore  at  least  one  root  lies 
between  2  and  3.     Hence  2  is  the  first  figure  of  a  root. 

2.  Find  the  integral  parts  of  all  the  roots  of  the  equation 
x3-Gx2+3x+9  =  0. 

3.  Find  the  first  figure  of  a  root  of  the  equation  x3—2x 
-  50  =  0. 

4.  Find  the  first  figure  of  a  root  of  the  equation  x*  —  2  ar8 
+  3x2-x-5  =  0. 

5.  Find  the  integral  part  of  a  root  of  the  equation  2  x4  +  x: 
-7x2-llx-4,  =  0. 

518.  To  find  the  superior  limit  of  the  positive  roots  of  an 
equation. 

Let  the  equation  be 

X=  xn  +  p  x"-1  +  q  .r"-2  + +  tx-  +  ux  +  v  =  0.        (1) 


a 


GENERAL   THEORY   OF  EQUATIONS.  301 

Let  r  be  the  numerical  value  of  the  greatest  negative 
coefficient,  and  xn~s  the  highest  power  of  x  which  has  a  nega- 
tive coefficient.  Then  the  first  s  terms  have  positive  coef- 
ficients. 

Now  Xwill  be  positive  when  x  is  positive,  provided 

x11  —  r  xn  ~ *  —  ),/_,_1  — —  r  x-  —  r  x  —  r  (2) 

is  positive ;  for,   since  r  is  the  numerically  greatest  negative 
coefficient,  and  all  terms  up  to  the  (s  +  l)th  are  positive,  Xis 
equal  to  (2)  plus  a, positive  quantity. 
We  may  write  (2) 

xn  —  r  (xn-s  +  xn-s-1  + +  x°-  +  x+l), 

or  (Art.  120),  x*-r  - — — ~ .  (3) 

x  —  1 

Then  AT  will  be  positive  when  (3)  is  positive.  But  if  x  is 
greater  than  unity,  (3)  is  evidently  greater  than 

xn  —  s  +  1 


X-l 


Therefore  X  will  be  positive  when  this  is  positive ;  or,  when 
(x  —  1)  x11  —  rxn-s  +  1  is  j>ositive  ;  or,  when  (x  —  1)  xs ~ x  —  r 
is  positive. 

But  (x—  1)  Xs-1  —  r  is  greater  than  (x  —  1)  (x  —  l)s_1  —  r 
or  (x  —  1)'  —  r ;  therefore  X  will  be  positive  when  (x  —  l)s  —  r 
is  positive  or  equal  to  zero  ;  or,  when  (x  —  l)s  =  r  or  >  r ; 
or,  when  x  —  l  —  \Jr  or  >  \/r;  or,  when  x  =  l-\-\jr  or  >1 
+  </r. 

That  is,  when  x  =  1  +  \J r  or  any  greater  value,  X  is  posi- 
tive, which  is  impossible,  as  it  must  equal  zero.  Hence  x 
must  be  less  than  l  +  tyr;  or,  1  +  $r  is  the  superior  limit  of 
the  positive  roots. 

519.  To  find  the  inferior  limit  of  the  negative  roots  of  an 
equation. 

By  changing  the  signs  of  the  alternate  terms  beginning 
with  the  second,  we  shall  obtain  an  equation  having  the 
same  roots  with  contrary  signs  (Art.  408). 


392  ALGEBRA. 

Then  evidently  the  superior  limit  of  the  positive  roots  of 
the  transformed  equation,  obtained  as  in  Art.  518,  will  by  a 
change  of  sign  become  the  inferior  limit  of  the  negative  roots 
of  the  given  equation. 

Note.  In  applying  the  principles  of  the  preceding  articles  to  determine 
the  limits  of  the  roots  of  an  equation,  the  absolute  term  must  be  taken  as 
the  coefficient  of  x'K 

520.     1.    Find  the  superior  limit  of  the  positive  roots  of 

x4  +  4  x3  - 19  x-  -  46  x  +  120  =  0. 

Here,  r  =  46,  and  n  —  s  =  2  ;  or,  as  n  =  4,  s  —  2.  Then  by 
Art.  518,  the  required  limit  is  1  +  y/46,  or  8  in  whole  num- 
bers. 

2.  Find  the  inferior  limit  of  the  negative  roots  of 

a-3 -a;2 -14.x +24  =  0.  (1) 

Changing  the  signs  of  the  alternate  terms  beginning  with 
the  second,  we  have 

x3  +  x2-Ux-2A  =  0.  (2) 

Here  r  =  24,  and  n  —  s  =  1,  or  s  =  2.  Then  the  superior 
limit  of  the  positive  roots  of  (2)  is  1  +  \/24 ;  therefore  the 
inferior  limit  of  the  negative  roots  of  (1)  is  —  (1  +  v/24). 

EXAMPLES. 

Find  the  superior  limits  of  the  positive  roots  of  the  follow- 
ing: 

3.  x*+2x3-13.r'1-Ux  +  24:=Q.    4.  a:4-15.T2+10,r+24=0. 
Find  the  inferior  limits  of  the  negative  roots  of  the  fellow- 


s' 


mg 


5.  x*-2x°--5x  +  (j  =  0.    6.  xx-5x5+ox-  +  5x  +  6  =  0. 

STURM'S  THEOREM. 

521.     To  determine  the  number  and  situation  of  the  real 
roots  of  an  equation. 


GENERAL  THEORY  OF  EQUATIONS.      393 

A  perfect  solution  of  this  difficult  problem  was  first  obtained 
by  Sturm,  in  1829.  As  the  theorem  determines  the  number 
of  real  roots,  the  number  of  imaginary  roots  also  becomes 
known  (Art.  487). 

522.     Let  X  denote  the  first  member  of 

Xn+21Xn-1+  qxn~2  + +  tX2  +  UX+  V  =  0, 

from  which  the  equal  roots  have  been  removed  (Art.  512). 

Let  Xx  denote  the  first  derivative  of  X  (Art.  511). 

Divide  AT  by  X1}  and  we  shall  obtain  a  quotient  Ql}  with  a 
remainder  of  a  lower  degree  than  Xv  Denote  this  remainder, 
with  its  signs  changed,  by  X2,  divide  Xx  by  X2,  and  so  on  ; 
the  operation  being  the  same  as  in  finding  the  greatest  com- 
mon divisor,  except  that  the  signs  of  every  remainder  must  be 
changed,  while  no  other  change  of  signs  is  admissible.  As 
the  equation  X  —  0  has  been  freed  from  equal  roots,  there  can 
be  no  common  divisor  of  Zand  J1;  and  the  last  remainder, 
X„ ,  will  be  independent  of  x. 

The  successive  operations  may  be  represented  by  the  fol- 
lowing equations : 

X  =  Xx  Qv  -  X2  (1) 

XI  =  X2Q2-X3  (2) 
X^X,Q,-X4                                (3) 


Xn  _  2  —  -<*«  - 1  V«  - 1      Xn 

The    expressions   X,    X1}    X2, XH  are    called    Sturm's 

Functions. 

STURM'S    THEOREM. 

523.  If  any  tiro  numbers,  a  and  b,  be  substituted/or  x  in 
Sturm's  Functions,  and  the  signs  noted,  the  difference  between 
the  number  of  variations  in  the  first  case  and  that  in  the 
second  is  equal  to  the  number  of  real  roots  of  the  given  equa- 
tion lying  between  a  and  b. 


394  ALGEBRA. 

The  demonstration  of  Sturm's  Theorem  depends  upon  the 
following  principles : 

(A).  Two  consecutive  functions  cannot  both  become  0  for 
the  same  value  of  x. 

For,  if  A\  =  0  and  X  =  0,  then  by  (2),  Art.  522,  X3  =  0 ; 
and  if  X2  =  0  and  X3  =  0,  by  (3),  X4  =  0;  and  so  on,  till 
Xn  =  0.  But  as  Xn  is  independent  of  x,  it  cannot  become 
0  for  any  value  of  x.  Hence  no  two  consecutive  functions  can 
become  zero  for  the  same  value  of  x. 

(B).  If  any  function,  except  X  and  Xn,  becomes  0  for  a 
■particular  value  of  x,  the  two  adjacent  functions  must  have 
opjjosite  signs. 

For,  if  X,  =  0,  we  have  by  (2),  Art.  522,  X1  =  —  Xs;  that 
is,  X1  and  Xs  must  have  opposite  signs,  for  by  (A)  neither  can 
be  equal  to  zero. 

(C).  When  any  function,  except  X  and  Xn,  changes  its  sign 
for  different  values  of  x,  the  number  of  variations  is  not 
affected. 

No  change  of  sign  can  take  place  in  any  one  of  Sturm's 
Functions  except  when  x  passes  through  a  value  which  re- 
duces that  function  to  zero. 

Now,  let  c  be  a  root  of  the  equation  X2  =  0;  d  and  e  quan- 
tities respectively  a  little  less  and  a  little  greater  than  c,  so 
taken  that  no  root  of  Xx  =  0  or  of  Xs  =  0  is  comprised  be- 
tween them.  Then,  as  x  changes  from  d  to  e,  no  change  of 
sign  takes  place  in  A^  or  X:i,  while  X2  reduces  to  zero  and 
may  change  sign.  And  as  by  (B),  when  X2  =  0,  Xl  and  Xz 
have  opposite  signs,  the  only  effect  of  a  change  in  the  sign  of 
X2  is  that  what  was  originally  a  permanence  and  a  variation 
is  now  a  variation  and  a  permanence  ;  that  is,  the  permanence 
and  variation  exchange  places.  Hence  a  change  in  the  sign 
of  X.,  does  not  affect  the  number  of  variations. 

As  Xn  is  independent  of  x,  it  can  never  change  sign  for  any 
value  of  x.     Therefore  a  change  in  the  number  of  variations 


GENERAL  THEORY  OF  EQUATIONS.      395 

can  be  caused  only  by  a  change   in   the  sign   of  the  given 
function  X. 

(D).  When  the  function  X  changes  its  sign  for  successive 
increasing  values  of  x,  the  number  of  variations  is  diminished 
by  one. 

Let  m  be  a  root  of  the  equation  X  =  0 ;  m  —  y  and  m  +  y 
quantities  respectively  a  little  less  and  a  little  greater  than 
7>i,  so  taken  that  no  root  of  X1  =  Q  is  comprised  between  them. 
Then,  as  x  changes  from  m  —  y  to  m  +  y,  no  change  of  sign 
takes  place  in  Xr ,  while  X  reduces  to  zero  and  changes  sign. 

Putting  in  +  y  in  place  of  x  in  X,  we  have 

O  +  y)n  +  P  (m  +  y)n~l  + +  u  (m  +  y)  +  v. 

B/eveloping  the  terms  by  the  Binomial  Theorem,  and  col- 
lecting terms  containing  like  powers  of  y,  we  have 

mn  +  p  mn  ~ 1  + +  um  +  v 

+  y  \_n  m"  _  l  +  p  (n  —  1)  nin  ~~ 2  + +  u\ 

+  terms  containing  y2,  ys, yn. 

Representing  the  coefficient  of  y,  which  we  observe  is  the 
value  of  X1  when  x  is  put  equal  to  m,  by  A ;  the  coefficient  of 
y1  by  B;  and  so  on,  we  have 

mn+pmn-1  + +  um  +  v  +  Ay  +  By-+ +  Kyn.   (1) 

But  as  x  =  m  reduces  X  to  0,  we  have 

mn  +  p  mn~1  + +  u  m  +  v  =  0. 

Hence  (1)  may  be  written 

Ay  +  Bf-+ +  Ky\  (2) 

Now  y  may  be  taken  so  small  that  the  sign  of  (2)  will  be 
the  same  as  the  sign  of  its  first  term.  That  is,  when  a?  is  a 
little  greater  than  m,  the  sign  of  X  is  the  same  as  the  sign 
of  Xl . 

Similarly,  by  substituting  in  —  y  for  x  in  X,  we  shall  arrive 
at  the  expression 

-Ay+By2-Cy3+ , 


396  ALGEBRA. 

where  as  before  y  may  be  taken  so  small  that  the  sign  of  the 
whole  expression  will  be  the  same  as  that  of  its  first  term. 
That  is.  when  a;  is  a  little  less  than  m,  the  sign  of  X  is  the 
reverse  of  the  sign  of  Xx. 

Thus  we  see  that  as  x  changes  from  m  —  y  to  m  +  y,  the 
signs  of  X  and  A^  are  different  before  x  equals  m,  and  alike 
afterwards.  Hence,  when  A  changes  its  sign  a  variation  is 
changed  into  a  permanence,  or  the  number  of  variations  is 
diminished  by  one. 

We  may  now  prove  Sturm's  Theorem ;  for  as  x  changes 
from  a  to  b,  supposing  a  less  than  b,  a  variation  is  changed  to 
a  permanence  each  time  that  X  reduces  to  0  and  changes  sign, 
and  only  then,  for  no  change  of  sign  in  any  of  the  other 
functions  can  affect  the  number  of  variations.  And  as  X 
reduces  to  zero  only  when  x  is  equal  to  some  root  of  the 
equation  X  —  0,  it  follows  that  the  number  of  variations  lost 
in  passing  from  a  to  l>  is  equal  to  the  number  of  real  roots  of 
the  equation  X  =  0  comprised  between  a  and  b. 

524.  When  —  oo  and  +  go  are  substituted  for  x,  or  when 
the  superior  limit  of  the  positive  roots  and  the  inferior 
limit  of  the  negative  roots  are  substituted  for  x,  the  whole 
number  of  real  roots  of  the  equation  A^  =  0  becomes  known. 

The  substitution  of  — go  and  0  will  give  the  whole  number 
of  negative  roots,  and  the  substitution  of  +  go  and  0  will  give 
the  whole  number  of  positive  roots.  If  the  roots  are  all  real, 
Descartes'  Kule  (Art.  507)  will  effect  the  same  object. 

The  substitution  of  various  numbers  for  x  will  show  be- 
tween what  numbers  the  roots  lie,  or  fix  the  limits  of  the 
roots. 

525.  A  and  A,  must  change  signs  alternately,  as  they  are 
always  unlike  in  sign  just  before  X  changes  sign  (Art.  f>L\'>>i 
(D)).  Hence,  when  the  roots  of  X  =  0  and  of  Aq  =  0  are  all 
real,  each  root  of  Aq  =  0  must  be  intermediate  in  value  be- 
tween two  roots  of  X=  0.  For  this  reason  the  first  derived 
equation  is  often  called  the  limiting  or  separating  equa- 
tion. 


GENERAL  THEORY  OF  EQUATIONS.      397 

526.  In  the  process  of  finding  X2,  X3}  etc.,  any  positive 
numerical  factors  may  be  omitted  or  introduced  at  pleasure,  as 
the  sign  of  the  result  is  not  affected  thereby.  In  this  way 
fractions  may  be  avoided. 

In  substituting  —  go  and  +  go,  the  first  term  of  each  func- 
tion determines  the  sign,  for  in  any  expression,  as 

axn  +  bxn~1  + +  /,•, 

where  x  may  be  made  as  great  as  we  please,  it  may  be  taken 
so  great  that  the  sign  of  the  whole  expression  will  be  the  same 
as  that  of  its  first  term. 

527.  1.  Determine  the  number  and  situation  of  the  real 
roots  of  the  equation 

x3  —  4  x2  —  x  +  4  =  0. 

Here,  the  first  derivative,  X1  =  3  x2  —  8  x  —  1.  Multiply- 
ing x3  —  4  x'1  —  x  +  4  by  3  so  as  to  make  its  first  term  divisi- 
ble by  3  x2, 

3x2-8x-l)3x3-12x2-   3 a: +  12  (a; 


3  a;3-    Sx2 


x 


-  4r-   2  a; +  12 

3 

-  6  a;2-    3  x  +  18  (-2 

-  6.r-  +  16a-  +    2 


19  a: +  16        .-.  X  =  19  x  -  16. 


3  x2  -      8  x  -       1 
19 

19  x  - 16 )  57  x2  - 152  x  -     19(3x 

57  x2  —    48  x 

-  104  x  -      19 
19 


1976  a;-    361  (-104 
1976  x  +  1664 

-2025  .-.  X  =  2025. 


398  ALGEBRA. 

Thus  we  have,  X  =  xz  —  4  x2  —  x  +  4 ;      X2  =  19  cc  —  10. 
Xj  ==  3  x-  -  8  x  -  1 ;  Z3  =  2025. 

The  last  step  of  the  division  may  he  omitted,  for  we  only- 
wish  the  sign  of  X3,  and  that  may  he  seen  by  inspection  when 
—  104  x  —  19  is  obtained. 

We  first  substitute  —  go  for  x  in  each  function,  and  obtain 
three  variations  of  sign  ;  similarly  +  go  gives  no  variation ; 
hence  the  three  roots  are  all  real.  Substituting  0,  we  have 
two  variations ;  comparing  this  with  the  former  results,  we 
see  that  one  root  is  negative  and  the  other  two  are  positive. 
The  same  result  could  have  been  obtained  by  Descartes'  Rule, 
as  all  the  roots  are  real.  We  now  substitute  various  numbers 
to  determine  the  limits  of  the  roots. 

The  table  presents  the  results  in  a  connected  form  : 


X 

*1 

^2 

*3 

When 

X  =  —  00, 

— 

+ 

— 

+ 

3  variations. 

Li 

x  =  —  2, 

— 

+ 

— 

+ 

3  variations. 

a 

x  =  —  1, 

0 

+ 

— 

+ 

it 

x  =  0, 

+ 

— 

— 

+ 

2  variations. 

n 

X  =  1, 

0 

— 

+ 

+ 

a 

x  =  2, 

— 

— 

+ 

+ 

1  variation. 

a 

%   ==  O} 

— 

+ 

+ 

+ 

1  variation. 

u 

x  =  4, 

0 

+ 

+ 

+ 

a 

OC  —  Oj 

+ 

+ 

+ 

+ 

no  variation. 

a 

X  =    GO, 

+ 

+ 

+ 

+ 

no  variation. 

Then  by  Sturm's  Theorem  we  know  that  there  is  one  root 
between  —2  and  0,  one  between  0  and  2,  and  one  bet  ween  3 
and  5.  In  fact,  as  X  ==  0  when  x  =  —  1,  1,  and  4,  these  are 
the  three  roots  of  the  equation. 

2.    Determine  the  number  and  situation  of  the  real  roots  of 


X 


x4  -  3  xs  +  3  x-  —  3  x  +  -  =  0 


Note.  In  substituting  the  various  numbers  to  determine  the  situation 
of  the  roots,  it  is  best  to  work  from  o  in  cither  direction,  stopping  when 
the  number  of  variations  is  the  same  as  has  been  previously  found  for 
+  oo  or  —  oo  ,  as  the  case  may  be. 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.    399 

Here  we  find  X,  =  4x3  -  9x2  +  6x  -  3  ;     Xa  =  —  <d2x  +  129  ; 
X  =  3  x2  +  18  x  -  31 ;        Xi  =  -  1163. 

Substituting  +  co  for  x,  we  obtain  one  variation  ;  similarly, 
0  gives  three  variations,  and  —  oo  gives  three  variations. 
Hence  there  are  only  two  real  roots,  both  of  which  are  posi- 
tive. We  then  substitute  values  of  x  from  0  upwards,  giving 
the  following  results : 


X 

^ 

x2 

Xs  X4 

When  x  =  0, 

+ 

— 

— 

+    — 

3  variations. 

■  "       x  =  l, 

+ 

— 

— 

+    - 

3  variations. 

"      x  =  2, 

+ 

+ 

+ 

—     — 

1  variation. 

"       x  =  co, 

+ 

+ 

+ 

—     — 

1  variation. 

Hence   there  are  two  roots  between  1  and  2 ;  and  as  the 
equation  has  four  roots,  there  must  be  two  imaginary  roots. 

EXAMPLES. 

Determine  the  number  and  situation  of  the  real  roots  of  the 
following  equations : 

3.  x*-x2-2x  +  l  =  Q.     6.   x*-2x9-5x2  +  10a;-3  =  0. 

4.  g*  —  7x+7  =  0.  7.   2x-4-3*3+  17x-2-3x-  +  15  =  0. 

5.  Xs  —  2  a -5  =  0.  8.  x*-  4  Xs -3  x  +  27  =  0. 


XLIIL  —  SOLUTION  OF  HIGHER  NUMERICAL 

EQUATIONS. 

528.  The  real  roots  of  the  higher  numerical  equations  in 
general  can  only  be  obtained  by  tentative  methods,  or  by 
methods  which  involve  approximation.  Cubic  and  biquadratic 
equations  may  be  considered  as  included  in  the  class  of  higher 


400  ALGEBEA. 

equations ;  for  their  general  solutions  are  complicated,  and 
only  of  limited  application.  No  general  solution  of  an  equa- 
tion of  a  degree  higher  than  the  fourth  can  he  ohtained. 


COMMENSURABLE    ROOTS. 

529.  A  commensurable  root  is  one  which  can  be  exactly 
expressed  as  an  integer  or  fraction  without  using  irrational 
quantities. 

An  incommensurable  root  is  one  which  can  only  he  ex- 
pressed approximately  by  means  of  a  decimal  fraction. 

530.  Any  equation  containing  fractional  coefficients  may 
be  transformed  into  another  whose  coefficients  are  entire,  that 
of  the  first  term  being  unity  (Art.  500),  and  such  an  equation 
cannot  have  a  root  equal  to  a  rational  fraction  (Art.  495)  ; 
hence,  to  find  all  commensurable  roots,  we  have  only  to  find 
all  integral  roots. 

531.  As  every  rational  root  of  an  equation  in  its  general 
form  is  a  divisor  of  the  last  term  (Art.  493),  to  find  the  com- 
mensurable roots  we  have  only  to  ascertain  by  trial  what  in- 
tegral divisors  of  the  absolute  term  are  roots  of  the  equation. 

The  trial  may  be  made  by  substituting  each  divisor,  both 
with  the  positive  and  the  negative  sign,  in  the  equation ;  or 
by  dividing  the  first  member  of  the  equation  by  the  unknown 
quantity  minus  the  supposed  root  (Art.  486).  In  substi- 
tuting very  small  numbers,  such  as  ±  1,  the  former  method 
may  be  most  convenient ;  but  when  an  actual  root  lias  once 
been  used,  the  latter  method  will  give  at  once  the  depressed 
equation,  which  may  be  used  in  obtaining  the  other  roots. 

532.  When  the  number  of  divisors  of  the  last  term  is 
large,  this  process  of  successive  trials  becomes  tedious,  and  a 
better  method,  known  as  the  Method  of  Divisors,  may  be 
adopted. 

If  a  is  a  root  of  the  equation 

x4  +pxi+qx2  +  tx  +  u  =  0, 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.     401 

then  a4  +  p  as  +  q  a-  +  t  a  +  u=  0. 

Transposing  and  dividing  by  a, 

-=  —  t  —  qa—pa2  —  a3,  (1) 

u 
whence  we  see  that  -  must  be  an  integer. 

a 

Equation  (1)  may  be  written 

— h  t  =  —  q  a  —p  a-  —  a*, 
cv 


U 

a 

a 


u 
Denoting  -  +  t  by  t',  and  dividing  by  a, 

—  —  q  —p  a  —  a2, 


t' 
whence  -  must  be  an  integer. 

a 


Proceeding  in  this  way,  we  see  that  if  a  is  a  root  of  the 
equation,  — \-  t  or  t',  — \-  q  or  q',  and \-  p  or  p1  must  be  in- 

Ct  Ct  Lt 

p' 

tegers,  and 1-1  must  equal  zero. 

Hence  the  following 

RULE. 

Divide  the  absolute  term  of  the  equation  by  one  of  its  inte- 
gral divisors,  ami  to  the  quotient  add  the  coefficient  of  x. 

Divide  this  sum  by  the  same  divisor,  and,  if  the  quotient  is 
an  integer,  add,  to  it  the  coefficient  ofx2. 

Proceed  in  the  same  manner  with  each  coefficient  in  regular 
order,  and,  if  the  divisor  is  a  root  of  the  equation,  each 
quotient  will  be  entire,  and  the  last  quotient  added  to  the 
coefficient  of  the  highest  poicer  ofx  will  equal  0. 

Equal  roots,  if  any,  should  be  removed  before  applying  the 
rule  ;  and  the  labor  may  often  be  diminished  by  obtaining  the 
superior  limit  to  the  positive  and  inferior  limit  to  the  nega- 


402  ALGEBRA. 

tive  roots  of  the  equation,  for  no  number  need  be  tried  which 
does  not  fall  between  these  limits. 

1.    Find  the  roots  of  the  equation 

xs  -  6  x2  +  27  x  -  38  =  0. 

By  Descartes'  Rule,  we  see  that  the  equation  has  no  nega- 
tive root;  and  the  only  positive  divisors  of  38  are  1,  2,  19, 
and  38.  By  substitution  we  see  that  1  is  not  a  root  of  the 
equation. 

Dividing  the  first  member  by  x  —  2,  we  obtain  x2  —  4  x  +  19 
as  a  quotient.  Hence  2  is  a  root,  and  the  depressed  equation 
is  x2  —  4  x  +  19  =  0,  from  which  we  obtain 


x  = 


4±y/16-76 

~~2~ 

as  the  remaining  roots.     Hence, 


2±v/-15 


x  =  2,  or  2  ±  y/  — 15,  Ans. 
2.    Find  the  roots  of  the  equation 

8  x*  -  4  xs  -  14  x2  +  x  +  3  =  0. 
We  may  write  the  equation 

A  9C  i  QC  0C  O  - 

*-2--^r  +  §  +  8=0- 

Proceeding  as  in  Art.  500,  we  see  that  the  multiplier  2  will 
remove  the  fractional  coefficients.     We  then  have  the  equation 

x4  —  xs  —  7  x2  +  x  +  6  =  0,  (1) 

whose  roots  are  twice  those  of  the  given  equation  (Art.  499). 

The  divisors  of  G  are  ±1,  ±2,  ±3,  and  ±  6. 

By  putting  x  equal  to  +  1  and  —  1  in  (1),  it  is  readily  seen 
that  both  are  roots  of  the  equation,  and  the  other  roots  can  be 
found  from  the  depressed  equation.  But  all  of  the  rational 
roots  may  be  obtained  by  the  rule. 


6, 

3, 

2, 

1, 

-1, 

-2, 

-3,-6 

1, 

2, 

3, 

6, 

-6, 

-3, 

-2,  -1 

2, 

3, 

4, 

("7 

-5, 

-2, 

-1,       0 

1, 

2, 

7, 

5, 

1, 

0 

-6, 

-$ 

o, 

9 

-6, 

-7 

-2, 

o, 

2, 

3 

-3, 

-1, 

1, 

2 

-1, 

-1, 

-1, 

-1 

o, 

o, 

o, 

0 

SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.    403 

It  is  customary  to  abridge  the  work  as  follows : 

Divisors, 
1st  Quotients, 
Adding  1, 
2d  Quotients, 
Adding  —  7, 
3d  Quotients, 
Adding  —  1, 
4th  Quotients, 
Adding  1, 

As  6,  2,  —  3,  and  —  6  give  fractional  quotients  at  different 
stages  of  the  operation,  they  cannot  be  roots  of  the  given 
equation,  and  are  rejected.  3,  1,  —  1,  and  —  2  give  entire 
quotients,  and  in  each  case  the  last  quotient  added  to  the 
coefficient  of  x4  gives  zero ;  hence  they  are  the  four  roots  of 

3  1       1 

equation  (1),  and  ^,  ^,  —  ^,  and  —  1  are  the  four  roots  of  the 

given  equation. 

EXAMPLES. 

Find  all  the  commensurable  roots  of  the  following  equa- 
tions, and  the  remaining  roots  when  possible  by  methods 
already  given  : 

3.  xs+Qx2+llx  +  Q  =  0.  8.  xs -7x2  +  36  =  0. 

4.  xs  +  3x2-4;x-12  =  0.  9.  x3-6x2  +  10a;  — 8  =  0. 

5.  a-4-4.r3-8x  +  32  =  0.  10.  x3-  6  x2  +  11  x-  6  =  0. 

6.  4a;8-16sc2-9a;  +  36  =  0.  11,  2x3-3x2+16x-2i  =  0. 

7.  x3-3x2  +  x  +  2  =  0.  12.  x5- 2 x3 -16  =  0. 

13.  xi-9x3  +  23x2-20x  +  15  =  0. 

14.  x4  +  x3  -  29  x2  -  9  x  +  180  =  0. 


404  ALGEBRA. 


RECURRING  OR  RECIPROCAL  EQUATIONS. 

533.  A  Recurring  Equation  is  one  in  which  the  coef- 
ficients of  any  two  terms  equally  distant  from  the  extremes 
of  the  first  member  are  equal. 

The  equal  coefficients  may  have  the  same  sign,  or  opposite 
signs;  hut  a  part  cannot  have  the  same  sign,  and  a  part 
opposite  signs,  in  the  same  equation.  Also,  if  the  degree  he 
even,  and  the  equal  coefficients  have  opposite  signs,  the  middle 
term  must  be  wanting.     Thus, 

a4  -  5  xz  +  6  x2  -  5  x  +  1  =  0, 

5  x5  -  51  x*  +  160  x3  - 160  x2  +  51  x  -  5  =  0, 

x6  —  x5  +  x*  —  x2  +  x  —  1  =  0, 

are  recurring  equations. 

534.  If  any  quantity  is  a  root  of  a  recurring  equation, 
the  reciprocal  of  that  quantity  is  also  a  root  of  the  same 
equation. 

Let  xn  +pxn~1  +  qxn~2  +  ...±  (...  +  </x2+i*z  +  l)=0      (1) 

be  the  equation.     Substitute  -  for  x  ;  then 

V 

y"      y'"1     yn~-  v  v     v      > 

Multiplying  each  term  by  yn, 

(i+py+.qf.+,..)±(...+  qyn-*+pyn-1+yn)=Q    (2) 

Now,  (1)  and  (2)  take  precisely  the  same  form  on  changing 

the  ±  sign  to  the  first  parenthesis  in  equation  (2),  and  hence 

they  must  have  the  same  roots.     Now,  if  a  is  a  root  of  (1),  as 

1    1 
v  —       -  must  be  a  root  of  (2) ;  but,  as  (1)  and  (2)  have  the 
x    a 

same  roots,  -  must  also  be  a  root  of  (1).     In  like  manner,  if 
a 

b  is  a  root  of  (1),  y  is  also  a  root  of  (1). 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.    405 

On  account  of  the  property  just  demonstrated,  recurring 
equations  are  also  called  reciprocal  equations  ;  the  former  term 
relating  to  their  coefficients,  and  the  latter  to  their  roots. 

535.  One  root  of  a  recurring  equation  of  an  odd  degree  is 
—  1  when  the  equal  coefficients  have  the  same  sign,  and  +1 
when  they  have  opposite  signs. 

A  recurring  equation  of  an  odd  degree,  as 

x*n  +  l  +px2m  +  qx2m-l  +   _        ±  (  +  q  tf  +  p  x  +  ^  _  Q  (3) 

has  an  even  number  of  terms,  and  may  he  written  in  one  of 
the  following  forms, 

(x2M  +  1  +  1)  +p  (x2m  +  x)  +  q  (x2"'-1  +  x")  +  =0, 

(X2m  +  1  —  1)  +p  (x2m  —  x)  +  q  (Xlm-1-X-)  +  ......  =  0. 

If  —  1  he  substituted  for  x  in  the  first  form,  or  +  1  in  the 
second,  the  first  member  will  become  0 ;  hence,  —  1  is  a  root 
of  the  first  and  +  1  a  root  of  the  second. 

If  equation  (3)  be  divided  by  x  ±  1,  both  forms  will  reduce 
to  the  following  form, 

X2m+px2"1-1  +  qx2m~2+ +  qx2+px  +  l  =  0,  (4) 

a  recurring  equation  of  an  even  degree  in  which  the  equal 
coefficients  have  the  same  sign.  Hence,  a  recurring  equation 
of  an  odd  degree  may  always  be  depressed  to  one  of  an  even 
degree. 

536.  Two  roots  of  a  recurring  equation  of  an  even  degree 
are  +  1  and  —  1  when  the  equal  coefficients  have  opposite  signs. 

Let 

x*>»+pX2m-1  +  qX2m-2+ -( +  qx2+px+  1)=0 

be  such  an  equation.  As  the  middle  term  must  be  wanting 
(Art.  533),  the  equation  may  be  written  in  the  form 

^™-l)+F(^»"!-l)  +  ?x2(f'»-4-l)-h =  0    (5) 

which  is  divisible  by  both  x  —  1  and  x  +  1,  or  by  x'2  —  1  (Art. 
120).     Hence,  both  +  1  and  —  1  are  roots  of  the  equation. 


406  ALGEBRA. 

If  equation  (5)  be  divided  by  x1  —  1,  it  will  be  depressed 
two  degrees,  and  become  a  recurring  equation  of  an  even 
degree,  in  which  the  equal  coefficients  have  the  same  sign 
(Art.  120).  Hence,  every  recurring  equation  may  be  de- 
pressed to  the  form  of  equation  (4),  Art.  535. 

537.  Every  recurring  equation  of  an  even  degree,  whose 
equal  coefficients  have  the  same  sign,  may  be  reduced  to  an 
equation  of  half  that  degree. 

Let 

a?m  +qi  x-"'-1  +  q  arm~2  + +  q  x~  +p  x  +  1  =  0 

be  such  an  equation.     Dividing  it  by  xm,  we  may  write  it 

(^  +  ^)  +  ^(^"-,  +  ^)+?(^-2  +  ^r-2)  +  ---=0     (6) 

the  middle  term  if  present  becoming  a  known  quantity. 
Put         x  +  -  =  y 

Then,  a2  +  -=-  =  y2  -  2 

x- 

x3+x^=y3~s(x+x)=i/3~sy 


xm  +  —  =  ym  —  m  ym  ~ 2  + 

Substituting  these  values  in  (6),  we  have  an  equation  of  the 
form 

y"l+2hym-l  +  qiym-*  + =  0. 

After  this  equation  is  solved,  we  can  immediately  find  x 

from  the  equation  x  +  -  =  ?/. 

x 

538.     It  thus  appears  that  any  recurring  equation  of  the 
(2  m  +  l)th  degree,  one  of  the  (2  m  +  2)th  degree  whose  equal 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.     407 

coefficients  have  opposite  signs,  and  one  of  the  2  with  degree 
whose  equal  coefficients  have  the  same  sign,  may  each  he 
reduced  to  an  equation  of  the  with  degree. 

EXAMPLES. 

1.    Given  x4  —  5  xs  +  6  x2  —  5  x  +  1  =  0,  to  find  x. 
Dividing  by  x2,    far  -\ — j  j  —  5  f  x  -\ — j  +  6  =  0. 

S instituting  y  for  x  -\ — ,  and  y2  —  2  for  x2  -\ — -9  we  have 

x  xl 

w2-2-5w+6  =  0. 

Whence,  y  =  4  or  1. 

If  y  =  4,  x  +  -  =  4,  or  x2  —  4  x  =  —  1 ; 

Whence,  x  =  2  ±  ^3. 

If  y  =  1,  x  +  -  =  1,  or  a;2  —  x  =  —  1 ; 


Whence,  x  = T 

Note.    That  2  —  y/3  and are  reciprocals  of  2  +  y/3  and  - — ^ 

1  2 

may  easily  be  shown  by  reducing  — — —  and  — - — ;=.  to  equivalent  frac- 

2  +  ^3  1  +  V—  3 

tions  with  rational  denominators  (Art.  279). 
Solve  the  following  equations  : 

2.  a-5  -  11  xl  +  17  x3  +  17  a-2  -  11  x  +  1  =  0. 

3.  a;5  +  2  x*  -  3  a;3  -  3  x2  +  2  x  +  1  =  0. 

4.  xG  —  x5  +  x4  —  x2  +  x  —  1  =  0. 

5.  a;3  +  px2+px  +  1  =  0. 

6.  6  x4  +  5  a.-3  -  38  x2  +  5  x  +  6  =  0. 

7.  5  x5  -  51  a,-4  +  160  a;3  -  160  a-2  +  51  x  -  5  =  0. 


408  ALGEBRA. 

8.  x4  +  5  x'  +  5  x  +  1  —  0. 

9.  x5  =  -  1,  or  x>  +  1  =  0.     (See  Art.  332.) 
10.   x5  -32  =  0.     (Let  cc  =  2  ?/.) 

CARDAN'S    METHOD   FOR  THE   SOLUTION   OF   CUBIC 

EQUATIONS. 

539.  In  order  to  solve  a  cubic  equation  by  Cardan's 
method,  it  must  first  be  transformed,  if  necessary,  into  another 
cubic  equation1  in  which  the  square  of  the  unknown  quantity 
shall  be  wanting. 

By  Art.  505,  this  may  be  done  by  substituting  for  x,  y 
minus  the  coefficient  of  x2  divided  by  3. 

540.  If  the  first  power  of  the  unknown  quantity  be  want- 
ing in  the  given  equation,  we  may  obtain  the  result  by  a 
simpler  method,  as  follows  : 

Let  x3  +  a  x2  +  c  =  0  be  such  an  equation. 

Substituting  -  for  x,  we  have 
V 

1        a 

—  +  —  +  c  =  0,  or  c  y%  +  a  y  +  1  =  0. 

541.  To  solve  a  cubic  equation  in  the  form  x3  +  p  x  +  q  =  0. 

P 
Put  x  =  z  —  =-,  and  the  equation  becomes 
Sz 

V2  V3  7)2 

or,  z3  -  £—.  +  q  =  0;  or,  27  z*  +  27  q  z3  -p3  =  0. 

2(  z3 

This  is   an  equation   in   the   quadratic   form,   and  may  be 

solved  by  the  method  of  Art.  313 ;  and  after  z  is  known,  x 

P 
may  be  found  directly  from  the  equation    x  =  «  —  —-. 

o  z 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.    409 
We  have  then  for  solving  cubic  equations  the  following 

RULE. 

If  necessary,  transform  the  equation  into  another  cubic 
equation  in  which  the  square  of  the  unknown  quantity  shall 
be  wanting  (Arts.  539  and  540). 

If  y  be  the  unknown  quantity  in  the  resulting  equation, 
substitute  for  it  z  minus  the  coefficient  of  y  divided  by  3z. 

EXAMPLES. 

1.    Solve  the  equation  x3  —  9  x  +  28  =  0. 

3 

Substituting  z  -\ —  for  x, 

27       97  27 

z3  +  9z  +  —  +^--9z  -  —  +  28  =  0, 

z        z6  z 

27 
or,  «3  +  tf  +  28  =  0;  or,  *c  +  28s3  =  -27. 

Solving  by  quadratics,     z3  =  —  1  or  —  27. 
Whence,  z  =  —  1  or  —  3. 

Uz  =  -1,  x  =  z  +  -  =  -l-3  =  -±. 

z 

If  z  =  -3,  x  =  -  3  - 1  =  -  4. 

Hence,  one  root  of  the  equation  is  —  4.  Dividing  the  first 
member  of  the  given  equation  by  x  +  4,  we  obtain  as  the  de- 
pressed equation, 

x2  —  4  x  +  7  =  0. 


Whence,  x  —  2  ±  y/—  3,  the  remaining  roots. 

2.    Solve  the  equation  xs  —  24  x2  —  24  x  —  25  =  0. 
Putting  x  =  y  +  8  (Art.  539),  we  obtain 
2/3  +  24^+192Z/  +  512-24y2-384?/-1536-24y-192-25=0, 
or,  y8  -  216  y  — 1241  =  0. 


410  ALGEBRA. 

72 
Putting  y  =  z  H — -,  we  have 

.      _,_        15552     373248      0._        15552      ,OM      _ 

zz  +  216s  h 1 5 216  « 1241  =  0, 

z  z*  z 

or,     z3  +  3'3^48  - 1241  =  0 ;  or,  sc  -  1241  z3  +  373248  =  0. 

Whence,  ss  =  729  or  512,  and  2  =  9  or  8. 

72  72 

Therefore,     y  =  9  +  -r  or  8  +  -^  =  17,  and  sc  =  y  +  8  =  25. 

9  o 

Hence,  one  root  of  the  equation  is  25.  Dividing  the  first 
member  of  the  given  equation  by  x  —  25,  we  have  as  the 
depressed  equation 

x2  +  x  +  1  =  0: 


- 1  ±  \J-  3     , 

Whence,  x  = - ,  the  remaining  roots. 

ii 

Solve  the  following  equations  : 

3.  x3-  6x  +  9  =  0.  6.  x3  +  9  x2-  21  x  +  11  =  0. 

4.  x3-  6 x2  +  57  x  -196=0.     7.  x3-2  x2  +  2x-  1  =  0. 

5.  :c3-4a;2-3x  +  18  =  0.  8.  x3-±x2  +  4a;-  3  =  0. 

9.   a-3-3x2  +  4  =  0. 
10.    Obtain  one  root  of  the  equation  x3  +  6  x  —  2  =  0. 

542.     In  the  cubic  equation  x3  +  px+q  =  0,  when  p  is 

—  p3      o2 
negative,  and     f,    >  =y  ,  Cardan's  method  involves  imaginary 

expressions ;  but  it  may  be  shown  in  that  case  that  the  three 
roots  of  the  equation  are  then  real  and  unequal. 

Thus,  in  solving  the  equation  x3  —  6  x  +  4  =  0. 

2 

Substituting  z  +  -  for  x,  we  have 
z 

z3  +  6  2  H \-  -j  —  Gz f-  4  =  0, 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.     411 
r,  g«  +  J*  +  4  =  0 ;  or,  zr>  +  4  z*  +  8  =  0. 


Whence,       zz  =  -2  ±  y/-  4,  or  -2  ±  2  y/-  1, 


or,  ft  =  y/_ 2  +  2y/-  1  or  ^-2-2^-1. 

It  may  be  proved  by  trial  that  1  +  y/—  1  is  the  cube  root  of 
_  2  +  2  yC3,  and  1  - y/^1  of  -  2  -  2  yCl.     Hence, 


=  l  +  y/-l  or  l-y/-l. 


If  z  =  1  +  y/-  1, 


X=Z+-=1+  y/— 1  + 


2         _2y/-l  +  2 

i  +  ^/=3_'  1  +  y'-i 


_  o 


Hence,  one  root  of  the  equation  is  2.  Dividing  the  first 
member  of  the  given  equation  by  x  —  2,  we  have  as  the  de- 
pressed equation 

x2  +  2  x  -  2  =  0. 

Whence,         £C  =  —  1  ±  y^,  the  remaining  roots. 

543  We  have  no  general  rule  for  the  extraction  of  the 
cube  root  of  a  binomial  surd ;  so  that  in  examples  like  that 
in  the  preceding  article,  unless  the  value  of  z  can  be  obtained 
by  inspection,  it  is  impossible  to  find  the  real  values  of  x  by 
Cardan's  method.  .  In  this  case,  the  real  values  of  x  can 
always  be  found  by  a  method  involving  Trigonometry. 

BIQUADRATIC  EQUATIONS. 

544.  General  solutions  of  biquadratic  equations  have  been 
obtained  by  Descartes,  Simpson,  Euler,  and  others.  Some  of 
them  require  the  second  term  of  the  equation  to  be  removed, 
while  others  do  not.  All  of  them  depend  upon  the  solution  of 
a  cubic  equation  by  Cardan's  method,  and  will  of  course  fail 
when  that  fails  (Art.  542).  They  are  practically  of  little 
value,  especially  as  numerical  equations  of  all  degrees  can  be 
readily  solved  by  methods  of  approximation. 


412  ALGEBRA. 

INCOMMENSURABLE   ROOTS. 

545.  If  a  higher  numerical  equation  is  found  to  contain 
no  commensurable  roots,  or  if,  after  removing  the  commen- 
surable roots,  the  depressed  equation  is  still  of  a  higher 
decree,  the  irrational  or  incommensurable  roots  must  next  be 
sought.  The  integral  parts  of  these  roots  may  be  found  by 
Sturm's  Theorem  or  by  Art.  517,  and  the  decimal  parts  by 
any  one  of  the  three  following  methods  of  approximation. 

HORNER'S  METHOD. 

546.  Suppose  a  root  of  the  equation 

x"  +j>x"~1  +  q  xn~2  + +  tx'2  +  ux  +  v  =  0         (1) 

is  found  to  lie  between  a  and  a  +  1.  Transform  the  equation 
into  another  whose  roots  shall  be  less  by  a  (Art.  502),  and  we 
shall  have  an  equation  in  the  form 

yn+p'Vn~1  +  Q'yn~2+ +  t'y°-  +  u'y  +  v'  =  0,     (2) 

one  of  whose  roots  is  less  than  1.  If  that  root  is  found  to  lie 
between  the  decimal  fractions  a'  tenths  and  a'  +  1  tenths, 
transform  equation  (2)  into  another  whose  roots  shall  be  less 
by  a'  tenths,  and  we  shall  have  an  equation  in  the  form 

s»+_p"2"-1  +  q"zn~2+  ......  +  t"z2  +  u"z  +  v"  =  0       (3) 

one  of  whose  roots  is  less  than  .1.     If  that  root  is  found  to 

lie  between  the  decimal  fractions  a"  hundredths  and  a"  + 1 

hundredths,  transform  equation  (3)  into  another  whose  roots 

shall  be  less  by  a"  hundredths ;  and  so  on. 

Thus  we  obtain 

x  =  a  +  a'  +  a"  + 

to  any  desired  degree  of  accuracy. 

As  y  and  z  in  equations  (2)  and  (3)  are  fractional,  their 
higher  powers  are  comparatively  small ;  hence  approximate 
values  of  y  and  z  may  be  found  by  considering  the  last  two 
terms  only,  from  which  we  have 

v'       ..  v" 

V  =  -77  and  *  ^  ~  TJ, ' 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.     413 

Thus  approximate  values  of  a',  a", may  be  found  in 

this  way,  and  with  greater  accuracy  the  smaller  they  become. 

Hence  a  positive  incommensurable  root  of  the  equation  may 
be  found  by  the  following 

RULE. 

Find  by  Sturm's  Theorem  the  initial  part  of 'the  root,  and 
transform  the  given,  equation  into  one  whose  roots  are  less  by 
th  is  initial  part. 

Divide  the  absolute  term  of  the  transformed  equation  by 
the  coefficient  of  the  first  power  of  the  unknown  quantity  for 
the  next  figure  of  the  root. 

Transform  this  last  equation  into  another  whose  roots  are 
less  by  the  figure  of  the  root  last  found,  divide  as  before  for 
the  next  figure  of  the  root  ;  and  so  on. 

547.  A  negative  root  may  be  found  by  changing  the 
signs  of  the  alternate  terms  of  the  equation  beginning  with 
the  second,  and  finding  the  corresponding  positive  root  of  the 
transformed  equation  (Art.  498).  This  hy  a  change  of  sign 
becomes  the-required  negative  root. 

548.  In  obtaining  the  approximate  value  of  any  one  of 

the  quantities  a1,  a", by  tbe  rule,  we  are  liable  to  get  too 

great  a  result ;  a  similar  case  occurs  in  extracting  the  square 
or  cube  root  of  a  number.  We  may  discover  such  an  error 
by  observing  the  signs  of  the  last  two  terms  of  the  next 
transformed  equation ;  for,  as  the  figures  of  the  root  as  ob- 
tained in  succession  are  to  be  added,  it  follows  that  a',  a", 

must  be  positive  quantities,  so  that  the  last  two  terms  of  the 
transformed  equation  must  be  of  opposite  sign.  We  then 
diminish  the  approximate  value  until  a  result  is  found  which 
satisfies  this  condition. 

549.  If  in  any  transformed  equation  the  coefficient  of  the 
first  power  of  the  unknown  quantity  should  be  zero,  the  next 
figure  of  the  root  may  be  obtained  by  dividing  the  absolute 
term  by  the  coefficient  of  the  square  of  the  unknown  quantity, 
and  taking  the  square  root  of  the  result. 


414  ALGEBRA. 

For,  if  in  equation  (2),  Art.  546,  u'  =  0,  we  have,  approxi- 
mately, 

t'  y2  +  v'  =  0,  whence  y  =  \/ 

We  proceed  in  a  similar  manner  if  any  numher  of  the 
coefficients  immediately  preceding  the  absolute  term  reduce 
to  zero. 

550.     1.    Solve  the  equation  x3  —  3  x2  —  2  x  +  5  —  0. 

By  Sturm's  Theorem,  the  equation  has  three  real  roots ; 
one  between  3  and  4,  another  between  1  and  2,  the  third 
between  —  1  and  —  2. 

To  find  the  first  root,  we  transform  the  equation  into 
another  whose  roots  are  less  by  3,  which  by  Art.  503  is 
effected  as  follows : 

Dividing  x3  —  3  x2  —  2  x  +  5  by  x  —  3,  we  have  x2  —  2  as  a 
quotient  and  —  1  as  a  remainder.  Dividing  x2  —  2  by  x  —  3, 
we  have  x  +  3  as  a  quotient  and  7  as  a  remainder.  Dividing 
x  +  3  by  x  —  3,  we  have  1  as  a  quotient  and  6  as  a  remain- 
der.    Hence  the  transformed  equation  is 

x3  +  6  x2  +  7  x  -  1  =  0, 

whose  roots  are  less  by  3  than  those  of  the  given  equation. 

Note.  The  operations  of  division  in  Horner's  Method  are  usually  per- 
formed by  a  method  known  as  Synthetic  Division.  For  example,  let  it  be 
required  to  divide  ofi  -  19  x  +  30  by  x  -  2. 


*3±0.r2-19a;  +  30 

T3  _  O  r2 


x  -2 


a;2  +  2x-15 


2  a;2 
2.-C2-    ix 


-15a; 
-15.Z  +  30 
0 


The  first  term  of  each  partial  product  may  be  omitted,  as  it  is  merely  a 
repetition  of  the  term  immediately  above.  Also  the  remaining  term  of 
each  partial  product  may  be  added  to  the  corresponding  term  of  the  divi- 
dend, provided  we  change  the  sign  of  the  second  term  of  the  divisor  before 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.    415 

multiplying.     Also   the   powers   of  x  may  be  omitted,  as  we  need  only 
consider  the  coefficients  in  order  to  obtain  the  remainder. 
The  work  now  stands 

l±0-19  +  30|l+2 


+  2  |  1  +  2-15 

+  2 
+   4 
-15 

-30 
0 

As  the  first  term  of  the  divisor  is  1,  it  is  usually  omitted,  and  the  first 
terms  of  the  dividends  constitute  the  quotient.  Raising  the  oblique 
columns  we  have  the  following  concise  form  : 

Dividend,  l±0-19  +  30|+2 

Partial  Products,  +2+    4-30 

Quotient,  1  +  2-15,+  0    Remainder. 

Here  we  use  only  the  second  term  of  the  divisor  with  its  sign  changed  ; 
each  term  of  the  quotient  is  the  sum  of  the  terms  in  the  vertical  column 
under  which  it  stands,  and  each  term  of  the  second  line  is  obtained  by 
multiplying  the  preceding  term  of  the  quotient  by  the  divisor  as  written. 

By  the  method  of  Synthetic  Division,  the  work  of  trans- 
forming the  given  equation  into  one  whose  roots  are  less  by 
3  stands  as  follows : 


1  - 

3 

— 

2 

+    5|  +3 

+ 

3 

0 

-   6 

0 

— 

2 

—    1,   1st  Remainder 

+ 

3 

+ 

9 

+ 

3 

+ 

7, 

2d  Remainder. 

+ 

3 

-f    6,    3d  Remainder. 

Thus  the  transformed  equation  is,  as  before, 

x3  +  6  x2  +  7  x  -  1  =  0.  (1) 

Dividing  1  by  7  we  obtain  .1  as  the  next  figure  of  the  root, 
and  we  proceed  to  transform  equation  (1)  into  another  whose 
roots  shall  be  less  by  .1. 


416 


ALGEBRA. 

6 

7 

-1 

.1 

.61 

.761 

6.1 

7.61 

-.239 

.1 

.62 

6.2 

8.23 

.1 

6.3 

Thus  the  transformed  equation  is 

xs  +  6.3  x2  +  8.23  x  -  .239  =  0, 
whence  by  dividing  .239  by  8.23  we  obtain  .02  for  the  next  root 
figure  ;  and  so  on.     Thus  the  first  root  is,  approximately,  3.12. 

Similarly,  the  second  root  may  be  shown  to  be  1.201  ap- 
proximately. 

By  Art.  547,  the  third  root  is  the  positive  root  of  the 
equation  x3  +  3  x~  —  2  x  —  5  —  0  with  its  sign  changed.  The 
successive  transformations  are  usually  written  in  connection 
as  in  the  following  form,  where  the  coefficients  of  the  different 

transformed    equations    are   indicated   by   (1),    (2),    (3),  

The  work  may  also  be  contracted  by  dropping  such  decimal 
figures  from  the  right  of  each  column  as  are  not  needed  for 
the  required  degree  of  accuracy. 

1 


3 

_  9 

4 

-5  |  1.33 

1 

2 

4 

2 

(1)173 

1 

5 

2.667 

5 

(1)7 

(2)-   .333 

1 

1.89 

(1)6 

8.89 

.3 

1.98 

6.3 

(2)  10.87 

.3 

6.6 

.3 

(2)  6.9 
Hence,  the  third  root  is  — 1.33  approximately. 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.     417 

EXAMPLES. 

Find  the  real  roots  of  the  following  equations : 

2.  ar3_2x-5  =  0.  5.   x3-  17  x2  +  54  x  -350  =  0. 

3.  a*  +  x3  -  500  =  0.         q    a;4 -4  a3 -3  a: +  27  =  0. 

4.  x3- 7  a; +7  =  0.  7.   x4  -  12  x1  +  12  x  -3  =  0. 

APPROXIMATION   BY  DOUBLE   POSITION. 

551.  Find  two  numbers,  a  and  b,  the  one  greater  and  the 
other  less  than  a  root  of  the  equation  (Arts.  517  or  521),  and 
suppose  a  to  he  nearer  the  root  than  b.  Substitute  them 
separately  for  x  in  the  given  equation,  and  let  A  and  B  repre- 
sent the  values  of  the  first  member  thus  obtained.  If  a  and  b 
were  the  true  roots,  A  and  B  would  each  be  0 ;  hence  the 
latter  may  be  considered  as  the  errors  which  result  from  sub- 
stituting a  and  b  for  x.  Although  not  strictly  correct,  yet, 
for  the  purpose  of  approximation,  we  may  assume  that 

A  :  B  =  x  —  a  :x  —  b 

Whence  (Art.  348),  A  —  B :  A  =  b  —  a  :  x  —  a 

or  (Art.  345),        A  —  B  :  b  —  a  =  A:x  —  a  (1) 

A(b-a) 

and,  x  —  a  =  — — 

A  —  B 

A(b-a) 
or,  x  =  a  +     A_B    . 

From  (1),  we  see  that,  approximately, 

As  the  difference  of  the  errors  is  to  the  difference  of  the  two 
assumed  numbers,  so  is  either  error  to  the  correction  of  its 
assumed  number. 

Adding  this  correction  when  its  assumed  number  is  too 
small,  or  subtracting  when  too  large,  we  obtain  a  nearer 
approximation   to    the   true   root.     This   result    and   another 


418  ALGEBRA. 

assumed  number  may  now  be  used  as  new  values  of  a  and  b, 
for  obtaining  a  still  nearer  approximation  ;  and  so  on. 

It  is  best  to  employ  two  assumed  quantities  tbat  sball  differ 
from  each  other  only  by  unity  in  the  last  figure  on  the  right. 
It  is  also  best  to  use  the  smaller  error. 

This  method  of  approximation  has  the  advantage  of  being 
applicable  to  equations  in  any  form.  It  may,  therefore,  be 
applied  to  radical  and  exponential  equations,  and  others  not 
reduced  to  the  general  form  (Art.  480). 

EXAMPLES. 

1.    Find  a  root  of  the  equation  xz  +  x-  +  x  —  100  =  0. 

When  4  and  5  are  substituted  for  x  in  the  equation,  the 
results  are  — 16  and  +  55,  respectively;  hence  a  =  4,  b  =  5, 
A  —  —  16,  and  B  =  55.  According  to  the  formula,'  the  first 
approximation  gives 

_  16  (5  -  4)      ,      16       ,  _ 
*  =  4+    -16-55    -4  +  7l  =  42+- 

As  the  true  root  is  greater  than  4.2,  we  now  assume  4.2 
and  4.3  as  a  and  b.  Substituting  these  values  for  x  in  the 
given  equation,  we  obtain  —  4.072  and  +  2.297  ;  therefore  4.3 
is  nearer  the  true  root  than  4.2. 

„  ,Q      2.297(4.3-4.2)  .2297 

Hence,      x  =  4.3-   ^  +  4_  =  4.3  -  -^ 

-4.3 -.036  =  4.264. 

Substituting  4.264  and  4.265  for  x,  and  stating  the  result 
in  the  form  of  a  proportion,  we  have 

.0276  +  .0365  :  .001  =  .0270  :  correction  of  4.264. 

Whence  the  correction  =  .00043+. 

Hence,         x  =  4.264  +  .00043  =  4.26443+,  Am. 

Find  one  root  of  each  of  the  following  equations : 


SOLUTION  OF  HIGHER  NUMERICAL  EQUATIONS.     419- 

2.  x3-  2  x-  50  =  0.  4.   a;3  +  8r+  Gx  —  75.9  =  0. 

3.  *3+10x2+5a;-260  =  0.    5.   x3+^-^-T  =  0. 

lb    4 

6.  x4  -  3  x2  -  75  x  -  10000  =  0. 

7.  a;5  +  2  x4  +  3  x3  +  4  a;2  +  5  x  -  54321  =  0. 

NEWTON'S   METHOD   OF  APPROXIMATION. 

552.  Find  two  numbers,  one  greater  and  the  other  less 
than  a  root  of  the  equation  (Arts.  517  or  521).  Let  a  he  one 
of  those  numbers,  the  nearest  to  the  root,  if  it  can  be  ascer- 
tained.    Substitute  a  +  y  for  x  in  the  given  equation  ;  then  y 

is  small,  and  by  omitting  y2,  y3, ,  a  value  of  y  is  obtained, 

which,  added  to  a,  gives  b,  a  closer  approximation  to  the  value 
of  x.  Now  substitute  b  +  z  for  x  in  the  given  equation,  and  a 
second  approximation  may  be  obtained  by  the  same  process  as 
before.  By  proceeding  in  this  way,  the  value  of  the  root  may 
be  obtained  to  any  required  degree  of  accuracy. 

The  assumed  value  of  x  should  be  nearer  to  one  root  than 
to  any  other,  in  order  to  secure  accuracy  in  the  approxima- 
tion. 

EXAMPLES. 

1.    Find  the  real  root  of  the  equation  x3  —  2  x  —  5  =  0. 

When  2  and  3  are  substituted  for  x  in  the  equation,  the  re- 
sults are  —  1  and  +  16  respectively  ;  hence  a  root  lies  between 
2  and  3,  and  near  to  2.  Substitute  2  +  y  for  x,  and  there 
results 

y3  +  6y"  +  10y-l  =  0. 
Whence,  approximately,  y  =  .1. 

Now  substitute  2.1  +  z  for  x,  and  there  results 
.061  +  11.23  2  + =  0. 


420  ALGEBRA. 

Whence,  approximately,  z  = '    i}0  =  —  .0054,  and 

x  =  2.1  -  -0054  =  2.0946,  nearly. 
Find  one  root  of  each  of  the  following  equations  : 
2.  a;8  —  3  x  -(-  1  =  0.         3.  xs  -  15  x-  +  03  x  -  50  =  0. 


ANSWERS    TO    EXAMPLES. 


In  the  following  collection  of  the  answers  to  the  examples  and  problems  given  in  the 
preceding  portion  of  the  text-book,  those  answers  are  omitted  which,  if  given,  would 
destroy  the  utility  of  the  problem. 


Art.  47;  pages  10  and  11. 

1.  93.         5.   408.          9.   5§.        13.   36.  17.  llf. 

2.  136.       6.   254.        10.   13$.      14.   48.  18.  9. 

3.  127.       7.   24.          11.   4.          15.   3.  19.  10. 

4.  156.       8.   310.         12.   If.        16.   4.  20.  76. 

Art.  60;  page  18. 

6.  Ua-9vip2.     7.  x.     8.  Sab-lcd.     10.  3mn2-2x2y. 
11.  39  a2 -2i  ab  +  5  h\  12.  -  «  +  3  c  +  2. 

13.  x-y+3m  +  3n.  14.  3a  +  3b  +  3c  +  3d.        15.  x. 
16.  n+ r.                  17.  6mn  —  ab  —  4  c  +  3x  +  3m2  —  4  p. 
18.  4  a  -  2  b  -  12  -  3  c  -  d  +  4  x2  -  18  m.  19.  6  a3. 
20.  14  six.       21.  7  a  b  +  7  (a  +  b).       22.  16  \/ y -  4  (a-  b). 

Art.  66;  page  21. 

6.  —  3a5+4ctf—  5  a  x.  7.  6  z  +  12  ?/  —  8  «  +  4. 

8.  _4a6c-14a--2y- 148.    9.  2  y/  a  -  4  y-  +  12  a .+  1. 
11.  14  x2  -Sj/2+  5ab-7.       12.  2  b -2  e.       13.  6  b  +  1. 

14.  4m^8»-r+3s.  15.  Qd -2  b  — 3  a -3c 
16.   5m2+9w3-71a;.               17.  2  b.  18.  a -6 -3  c. 


422  ■       ALGEBRA. 

Art.  74;   page  24. 

4.  a  —  b  +  c  +  cl  —  e.     5.  2 a  +  2.      6.  x  —  y.      7.  a  —  3b  +  c. 

8.  5wr-6»-4(i.  9.  6?»  —  3%.  10.  4x  +  2y. 

11.  —35  —  7c.        12.  a  — c.         13.  9a+l.         14.  6  m  +  2. 

Art.  86;  pages  29  and  30. 

3.  6a3-16a2?/+6ay2  +  4?/3.  4.  a4  +  4a  +  3.  5.  a2-b2 
+  2bc-c\  6.  -0a2+16a&-8£2.  7.  68-a8.  8.  o4-«. 
9.  30  a3  -  43  a2  b  +  39  a  b2  -  20  b3.  10.  6  a4  +  13  x3  -  70  a3 
+  71a-20.  11.  -a5- 37a;2  +  70 x -50.  12.  -Gar5- 25 a4 
+  7a3  +  81a2  +  3a-28.  13.  2a5b2-3a4  b3-7  a3b4  +  4:a2bs. 
14.  4x2m+1  y3  —  1G  xm+6  yn  +  1  +  12  x5  y2n~\  15.  12  a;6 +7  a;4 
+  5  x3  +  10  x  —  4.  16.  m5  +  re5.  17.  a5  —  5  a4  6  +  10  a3  b2 
-10  a2b3  + 5  ab4-  b\ 

Art  87;  page  30. 

2.  6  a2  +  11  a  J  +  4  62.  3.  a5  +  x5.  4.  a8  -  2  a4  a4  +  x8. 
5.  2  am  +  1  -  2  an  +  1  -  am  +  n  +  a2n.  6.  1  -  a8.  7.  a3  +  3  a2  a 
-  10  a  x2 - 24 x3.     8.  a5 -5  a4  +  10  a3 -10  a2  +  5  a-  1. 

Art.  101;  pages  37  and  38. 

3.  a  x - 2.      4.  3 b2 -  4  a2.      5.  4 «2 - 3  b2.      6.  3«4  +  3a3i 

+  3a262  +  3«63  +  364.     7.  a2-ax  +  x2+-^—.    8.  x3-x2y 

a  +  x 

+  xy2-y3  +  —¥—.    9.  2  x2  -  7  a  -  8.     10.  5  x2  -  4  a  +  3. 
a;  +  y 

11.  x2  -  2  x  -  3.  12.  a4  +  x3  y  +  a-2  ?/2  +  x  y3  +  y\ 

13.  3a3-2a2  +  a-5.         14.  2  a-3  -  a-  +  1.         15.  a-b+c. 

16.  a2— 3  a—  y.  17.  x  +  //.  18.  an  —  bm  +  cr. 

19.  l  +  2a  +  2a2  +  2a3+  ...        20.  a-ax  +  ax2-ax3+  ... 


ANSWERS   TO   EXAMPLES.  423 

21.  a*-an  +  aib2-ab*  +  bi.  22.  2  a3-  2  a2-  3  a  -  2. 

23.  -  x2  -  2  x  -  4.       24.  a3  -  a:  +  2.       25.  2  a2  -  a  b  +  2  &2. 

Art.  107;  page  40. 

23.  1  -  a2  +  2  a  b  -  b2.  24.  a2  -  b2  -  2  b  c  -  c2. 

25.  a2-2ab  +  b2-c2.  26.  c2- a2  + 2  «6 -£2. 

27.  a2  +  2«i  +  i'2-c2  +  2^-  c/2.  28.  a2  -  2  a  5  +  52 

-c2+2crf-£  29.  a2  +  2  a  &  +  &2  -  c2  -  2  c  d  -  d2. 

Art.  115;  page  42. 

3.  (a  +  x)(b  +  y).      5.  (x  +  2)(x-  y).     7.  (x2  -  y2)  (m  -  n). 

4.  (a-m)(c  +  tf).      6.   (a-b)(a2  +  b'2).     8.  (a:  +  1)  (x2  +  1). 
9.  (3  a; +  2)  (2  a;2 -3).  12.  (ab-cd)  {ac  +  b  d). 

10.  (2 a;  - 3 y)  (4  c  +  d).    ,       13.  (m2 x-ny)  (n2 x  -  m y). 

11.  (2-7m2j(3»-4m).       14.  (4ww-7a;y)(3a&  +  5crf)- 

Art.  117;  page  45. 

9.  (a  +  b  +  c  +  d)  (a  +  b  —  c  —  d). 
10.   (a  —  c  +  b)  (a  —  c  —  b).         11.   (m  +  x—y)  (m  —  x  +  y). 
12.   (x  —  m  +  y  —  n)  (x  —  m  —  y+  n). 

16.  (x  +  y  +  2)(x  +  y-2).      18.   (3c  +  d+l)(3c  +  d-l). 

17.  (a  +  b-c)(a-b  +  c).        19.   (3  +  x2-2y)  (3-x2  +  2y). 

20.  (2  a  —  5  +  3  d)  (2  a  -  J  -  3  d). 

21.  (2  m2  +  2  b  -  1)  (2  m2  -  2  b  +  1). 

22.  (a  —  m  +  b  -f-  n)  (a  —  vi  —  b  —  n). 

23.  (a  +  m  +  b  —  n)  (a  +  m  —  b  +  n). 

24.  (x  —  c  +  y  —  d)  (x  —  c  —  y  +  d). 

Art  118;  page  49. 

25.  (x2  -  24)  (x2  -  5).  27.  (x  y3  +  12)  (x  y3  -  10). 

26.  (c3  +  11)  (c3  +  1).  28.   (a  b2  -  16)  (a  b2  +  9). 


424        •  ALGEBRA. 

29.  (x  +  20  n)  (x  +  5  n).  32.   (x  +  y  -  5)  (x  +  y  —  2). 

30.  (m2  +  11  n1)  O'2  -  6  ir).      33.  (x  -  8  y2  g)  (aj  +  6  y2 s). 

31.  (a  _  5  _  4)  (a  _  5  +  1).       34.  (m  +  n  +  2)  (m  +  ra  -  1). 

Art.  121 ;  page  53. 

3.  Sab  (a  -'r  2)2.  7.  3  a2  (a  -  5)  (a  -  2). 

4.  5  x  y2  (3  x  -  4  y2)2.  8.  2  c  m  (c  +  7)  (c  -  3). 

5.  2ay(3a;  +  y)(3a;  — y).  9.  a;  y  (m  — 6)  (m  +  2). 

6.  a;  (x  +  7)  (a;  +  1).  10.  ±ab(2a  +  b)  (±a2-2ab  +  b2) 

11.  (n  -  1)  (>r  +  »  +  1)  O6  +  w3  +  1). 

12.  (x2  +  y2)(x-t-2/)(^-2/). 

13.  (cc4  +  m4)  (a;2  +  m2)  (x  +  m)  (x  —  m). 

14.  (m  +  w)  (??i  —  w)  (m2  +  m  w  +  ?z2)  (??i2  —  mn  +  ri2). 

15.  (a  +  c)  (a2  -  a  c  +  e2)  (a6  -  a3  c3  +  c6). 

16.  (2  a  +  1)  (2  «  -  1)  (4  a2  +  2  a  +  1)  (4  a2-  2  a  +  1). 

Art.  125 ;  pages  55  and  56. 

Z.  ax.  6.  x  +  7.  9.  x  (x  —  1).         12.  2  a:  +  5. 

4.  m  +  n.        7.  2  a;  —  3.        10.  a  —  2  &.  13.  m(j;-  1). 

5.  x2  +  1.       8.  3  a  -  4.        11.  x  +  6.  14.  4  a?  -  1. 

Art.  126 ;  page  61. 

6.  2  x  +  3.  10.  2  a;  -  5.  14.  x2  +  x  +  l. 

7.  8a;  — 7.  11.  5x  +  3.  15.  a  — a;. 

8.  x  - 1.  12.  jb  +  2.  16.  x2  -  2. 

9.  3  a;  +  4.  13.  2  a;  -  1.  17.  2  (x  +  y). 

18.  2«-3x.  19.  3 a; +  2. 

Art.  130;  page  63. 

2.  120a462c.  4.  36  a&b\  6.  840  a2  c2  d3. 

3.  30xV«8.  5.  480  m3  rt2  x2  t/2.  7.  252  a8  y3  s8. 

8.  1080  a2  b2  c3  d\  9.  168  m  n2  xz  ys. 


ANSWERS   TO  EXAMPLES.  425 

Art.  131;   pages  63  and  64. 

2.  ax(x-\-a)(x—a)(x-+ax+a2).  7.  ax  (x  —  3)  (x—  7)  (x  +  8). 

3.  12 abc (a  +  b)(a-b).  8.  (2x+l)(2x-l)%4:x2+2x+l). 

4.  «(cc  +  l)(a;-l)(x2— rc  +  1).  9.  3  ab  (x  —  y)2  (a  —  b). 

5.  24(l  +  cc)(l-x)(l  +  a:2).      10.  2az;2(3;c  +  2)2(9a;2-6x+4). 

6.  (x+l)(a;-2)(cc+3)(*+4).    11.  (x-l)(x-3) (x  +  4)(x-5). 

12.   (x  +  y  +  z)  (x  +  y  —  z)  (x  —  y  +  z). 

Art.  132;  page  65. 

2.  (3x-4)(4:r-5)(2a;  +  7).    4.  (a2-2a-2)(a  +  3)(2a-l). 

3.  (4a;+l)(2a;+7)(3a;-8).    5.  (2x  +  3)(x2-x  +  l){x2+x-2). 

6.  (a  —  b)(a2-ab  +  b2)(a2  +  2ab  +  b2). 

7.  a  x  (x  +  1)  (x2  —  x  —  1)  (x2  +  x  +  1). 

8.  x  (x  -  5)  (2  x2 - x  -2)  (3 x2  +  x  -  1). 

If  the  above  expressions  are  expanded,  the  answers  take  the 
following  forms : 

2.  24  z3  + 22  a2 -177  a; +  140.     4.  2«4  +  a3-17«2-4«  +  6. 

3.  24  z3  +  26  z2- 219  a; -56.        5.  2x5  +  3xi-±x3  +  5x-Q. 

6.  a5-a3b2  +  a2b3-b\ 

7.  a xG  +  ax5  — ax4  —  3  ax3  — 3  ax2  — ax. 

8.  6ic6-31a;c-4x4  +  44a;3  +  7a;2-10a;. 

Art.  148  ;  page  71. 

14-  *£■■  18- 

o  +  2  c 
a  (2 +  3  n) 
1D*   b{2-3n)  •  19' 

16.  ±*--J*y+v\      20. 

m  +  9  3  y  —  5 


10. 

cd 
3  xy' 

11. 

xa 
2?' 

12. 

sc  — 5 

#  +  7' 

13. 

m  —  2 

2 

x2y 

5 

—  X 

2  +  x 

X 

c 

(J-x) 
-d 

c 

III 

+  d' 

0 

.  —  nr 

m2  —  n 


426  ALGEBEA. 

Art.  149;  page  72. 

2.3* 


3. 


4  a;  +  1 

5a +  7 
a-2  ' 

10. 


4. 

m  —  1 

6  m  —  5 

5. 

x  +  2 
x  —  3 " 

x' 

2 

3a;  +  l 

x'z  —  x  +  3 

Art.  150;   page  73. 


6. 

3  a;  —  2 

8    2a;  +  5 
2  x  -7 

a;  +  3  * 

7. 

2x-3 

_    C  «  —  1 

"•  7= » 

5a  — 7 

2a;-5* 

,  1      2x2-a;-2 

U<  2  a2 

+  3a;  +  l* 

3 

a  - 

a* 

4. 

a;2 

—  X 

!/  +  */• 

6. 

a;2 
3" 

a; 
~~3 

7 
+  3" 

2 
a;" 

7. 

a 
2b' 

3      2b 

2+  a  ' 

K   2x      3 
o 


5        5      5a; 
23 


8.  2a;  +  6  + 


9.  x2  +  a;  +  l.      10.  2  +  77-^-    — ^-,      11.  a;-24 

12.  jc 


x  —  3 
2a;-4 

2xz  —  x+1  ,xz+x — 1 

a;  — 2 
2a;2-3a;  +  3' 

Art.  151;  page  74. 

_     (x  —  1)2  _    an  +  b-—cd  56.r  — 4?i2 —  5a 

2.  =— .  o.  .  4.  - . 

x  —  6  n  o 

(a;  +  l)2  .    2ab  „    a2+2b2  Q    a3  +  b3 

0.  .  b.  j.  7.  — s .  o.  — . 

x  a+b  2  a  a  —  b 

Q   6a;2-7a;-l  2b2  ,,    a;3-2a;2-3a; 

"•   ^ : n •  I".    — ■ -.  11.    „ . 

2a;  — 1  a+&  x  —  2 

Art.  152;  pages  76  and  77. 

27  a  b    16  a  c    30  6  n  3  a;2  ?/     2  a-?/,?    7  y  z2 

3'  772~'  "T^7  ~72~"  30    '      30    '      30    ' 

18  ?/a2    16  x2  z2    15  a;2  y2  ' 
12  xyz*  12xij  z'  12  xyz' 
40c2-10c     18i2-12  6      25  «a  _2^_     3_o^    4  a  a;2 

30  a&c    '       30  abc     '  30  a  be'       '  a*x8'   a8  a;8'    a8  a;3  ' 


8. 


ANSWERS  TO  EXAMPLES.  427 

100  a  y  z3       Aob  x3  z       8Acxys  —  12mx  y2 
120  x1  ?fz2'  120  x2y2z2>  '       120  x2y2z2         * 


(a+b)  {a2  +  U1)      (a  -  b)  (a2  +  b2)      a2  -  b2 


10. 


a4-64  at-b*  '   at-W 

x2  —  9  x2-l  x2-4 


(x-l)(x-2)(x-3)'  (x-l)(x-2)(x-3)'  (x-l)(x-2)(x-3)' 


2  a  (a +  2)  3  b  (a  — 2)  4  c  (a  +  3) 


12. 


0-2)(a+2)0+3)'  0-2)0+2)0+3)'  (a-2)(a+2)(a+3)' 
x3+2x2-\-2x  +  l       x2  +  x  +  l  x  +  1 


(x  +  1)  {x3- 1)    '0  +  1)  (Xs-  1)'  {x  +  1)  (a3- 1)' 
13  6  a2  b2  3  b  (m2 - n2)  2  a  (a2-  b2) 


6ab(a  —  b)(m  +  ?iy  6ab(a—b)(m  +  n)'  6ab(a—b)(m  +  ii)' 

3 Q+l)   2Q-1)    2-a  1-x  x2-x-2       3 

'     a»-l'    a2-l'a2-l'         "*'   l-x»     1-jb2   >1=?' 

17  c2-^2  Q-l)Q  +  &)  Q-6)2 

'  O2-  62)  (c  -  d)'  O2-*2)  (e-d)'  (a2-b2)  0  -  d)' 

Art.  153;  page  78. 

2    (a~by        3    g2+9«+8  9  m2 - 4 

a*  —  b2'         '  x2+5x  —  24'  "6m2— 19m  +  10" 

4  02+«&  +  62)  1_^ 

a3  —  &3  1  —  x 

Art.  154;  pages  80  to  82. 

4    12  a; +  7  6a  +  5&  a  +  3  3  m2  n2  -  4 

36      '        '     10  a2 62  '  24    •  6m2-»3     * 

8    5  &*  +  4  «'        q    5a  +  6       in    4,ab-b-±a3  1 

"      120  a  6     "      *       24     •      iUl  12«3T~  ~*      1L  15" 

12    —      is    3a;~2  1  45ctf+6acd— 3ahd— 2abc 

•42'    "■     18  x2'   14l_60-   15-~         ~48^X~ 

17.  ? 18    1 19    2Q^  +  &2) 

6  +  x-x2'  '  x2+15x+56'  a2-b2    ' 


428  ALGEBKA. 

■Oi     *"        «l£±|.    22..  4^.    23.        (*  +  2y 


1_^'     —   a-6*          -a4-^'  •  (.x  +  l)(a;3-l) 

13-18*                        _J_  a2-14a  +  l 

^  (z  +  lXa^Xz-S)*       0,&-«'  '       6(^-1) 

28.  57_?-_.    29.  44-    30.  0.     31. 


9  a:- a;3'        '  »*— 1"         '     '        '       (»-2)(jb-3)(»-4) 
Art.  155;  pages  83  to  85. 


2. 

a5bsc 
m*  n3 d' 

3. 

12  a4  b  x 
35  h5  m  ' 

4     ^ 

7. 

/  •    8- 

4  a.'  y 

a3. 

911 

o 

m 
4 

w2               3  a; 

a;  - 

-1  10* 
-2  *       "■     3 

13. 

b(a-b) 
x(a  +  b) 

14. 

a  —  b 
a2     ' 

I*'1-'. 

a; 

ia  x*-x-2Q 
16' 

17. 

a  (x-2) 
a  +  1 

18. 

X 

x-2' 

19.  "+*• 

.     or 

20.  aj2  +  5f  +  6 

ST 

21.  -^o .        22.  x-  +  xy.       23.  1.       24.  2.        25.  -J£-j. 

ic  +  2  %  ~r  y 


.    91  ra2 
o. 


6rc2 

2 


4. 


Art.  156 

;  page  86. 

5  mx 

7.!. 

X 

9.»  +  1     ' 

a  +  5 

,   3  0-4) 

8    3.t- 2?/ 

10.  aj. 

8  b3  m  ri1  xs  x  +  y 


Art.  157;  pages  88  and  89. 


a 


_     acn  +  bn         „    3m— n      „    Ay—±x  +  2a 

— — .      o.  —        — .      b.  — x .     «. ^j • 

bm  +  b?i  cnx — cm  ox  oi 

x2y3+l 


8.  x  —  1.         9.  a;2-rK  +  l.  10.  a  +  b.  11. 


xif  —  ly* 


,n    a— J         10    «— 4       ...  1(.         4  R        aft 

12.  -T7-.        13. ^.      14.  aj.      15.  k—  -Q-       lb-     ■>  ,  7,2 ■ 

a-  b  x  +  6  3  a; +  3  a"  +  0 


ANSWERS   TO  EXAMPLES.  429 


17    1  18         ww(m-w)8  19_    s-g 

m*  +  m1  ri1  +  w4'  x  +  2  a 


Art.  175;   pages  94  and  95. 

2.  aenx—bcen=bdnx—bem.       3.  6  bx  —  8a2  =  3  —  2a6a. 

4.  bdex— adex+bcex— abd=0.      5.  \2x-\-hx  —  6x— 1320- 

6.  9z-12a=10a:+24-46.     7.  28a:-4a;+560=14a;+7a;+728. 

8.  ±ax-6c-5asx  +  2a3bd=0.  9.  10x-32a;-312=21-52a;. 
11.  3  a  —  2  a  —  2  a;  =  45.  12.  a  6  a;  +  b1  —  ex  —  d  =  a  c. 

13,  3-3z-2-2x  =  0.  14.  6ar+3z-6ar+  18-4a;-2=0. 
15.  3a._3_2a--2-5x=0.     16.  6x  +  6-15aj  +  45-20a;-10=0. 


Art. 

177;  pages 

97  to  102. 

4. 

3. 

13.  1. 

22.  72. 

31.  5. 

41.  - 1&. 

5. 

7. 

14.  2. 

23.  60. 

32.  -5. 

42.  4£. 

6. 

-1. 

15.  2. 

24.  10. 

33.  4. 

43.  ltV 

7. 

16.  -4. 

25.  -2* 

34.  -5. 

44.  0. 

8. 

1. 

17.  2. 

26.  56. 

35.  -2. 

45.  H. 

9. 

3 

< 

18.  1.     > 

27.  ?. 

33.  f . 

46.-^. 

11. 

3 

—  2' 

20.  3. 

29.  -2. 

37.  -1. 

47.  1. 

12. 

0. 

21.  5. 

ao.J. 

40.  -7. 

50.  f  -* 

2a+b 

51. 

a2 

a2- 

+  4a0.      52 
3a+2 

!.  2&.      53. 

If         *I 

Abc  +  a* 
a?—b  +  16c 

55. 

2  a2 
36 

,      57.  ^. 

0 

59.         \ 

a  +  2 

i-        61--^ 

64.  -3. 

56. 

a. 

r 

58.  12  a3. 

60.  aft. 

63.  2. 

65.  50. 

66B5- 

67.  5. 

68.  0. 

430  ALGEBEA. 

Art.  182 ;  pages  108  to  113. 
10.  Horse,  $224;  chaise,  $112.  11.  37.  12.  10  and  7. 

13.  18  and  2.  14.  58J  and  4l£.  15.  A,  40 ;  B,  20. 

16.  A,  60 ;  B,  15.  17.  If.  18.  ft.  19.  23£. 

20.  84.  21.  36.  22.  Oxen,  12 ;  cows,  24. 

23.  Wife,  $864;  daughter,  $288;  son,  $144. 
24.  Worked,  20  ;  absent,  16.      25.  Horse,  $  126  ;  saddle,  $  12. 

26.  Infantry,  2450 ;  cavalry,  196 ;  artillery,  98. 
27.  144  sq.  yds.  28.  Water,  1540 ;  foot,  880  ;  horse,  616. 

29.  $  1728.  30.  $  2000  at  6  p.c. ;  $  1200  at  5  p.c.  31.  7. 
32.  31.  33.  $24.  34.  $100.  35.    142857. 

36.  A,  $  466| ;  B,  $  533J.  37.  2  dollars,  20  dimes,  4  cents. 
38.  $2.75.  39.  Men,  $25;  women,  $21.  40.  23  and  18. 
41.  48  minutes.  42.  12121  men ;  110  on  a  side  at  first. 

43.  5r\  minutes  after  7.  44.  43T7T  minutes  after  2. 

45.  27f\  minutes  after  5.  46.  29  and  14. 

47.  3377  ounces  of  gold ;  783  ounces  of  silver.  48.  $  2000. 

49.  30  bushels  at  9  shillings ;  10  at  13  shillings.  50.  10  a.m. 
51.  $  1280.  52.  21  ft  minutes,  or  54T<Y  minutes  after  7. 

53.  27^T  minutes  after  4.  54.  23*1  miles. 

55.  Greyhound,  72  ;  fox,  108.         56.  1  minute,  If  §f  seconds. 

Art.  192 ;  pages  120  to  123. 

3.  x  =  4,  y  =  3.  9.  x=-2,  y=10.    15.  x  =  12,  y  =  18. 

4.  s  =  5,  y  =  -2.     10.  jk==12,  y  =  S.      16.  a;  =  35,  y  =  -10. 

5.  x  =  7,  y  =  5.         11.  x=~2,  y=-10.  17.  x  =  -  28,  y  =  21. 

6.  x  =  -  8,  y  =  2.     12.  x  =  10,  y  =  5.      18.  x  =  A,y  =  .1. 

7.  x  =  5,  y  =  7.         13.  x  =  7,  y  =  11.      19.  x  =  l£,  y  =  3f . 

8.  x=-8,  y=-12.  14.  a>=ll,  y=-9.    20.  x  =  3,  y  =  -2. 
21            dm  —  bn        _an  —  c  m  n  /  +  ri  r 

ad  — be  ad  — be  ~  mri  +  m'  n ' 


ANSWERS  TO  EXAMPLES.  431 

m'r—mr'       nn  ac(bvi  +  dv)  bd(cn-am) 

V  = •      "3.  X  = ^-= ; -,  y  = ^ : -. 

u      m  n'  +  w!  n  ad  +  b  c  ad  +  b  c 

11  25  5* 

24.  x=—,  y=-^.    25.  x  =  60,  y  =  40.     26.  x  =  —  ,  y=». 
2  a  2,  a  Ob 

27.  x  =  lft,  y  =  4TV    28.  x  =  -6,  y=-5.    30.  #  =  4,  y=2, 

be  —  ad 

bn  —  d  vi  ' 


31.  x  =  —  5,  y=3.     32.  x  =  -2,  y  —  —  l.     33. ■  x  = 


b  c  —  a  d        „.  3  2  oe  1  1 

y  = •      34.33  =  ^77,2/  =  —^.      35.  x=-,y  =  — 

cm  — an  a~  b  a  tr  n  vi 


Art.  194;  pages  126  and  127. 

3.  x  =  23,  y  =  6,  g  =  24.  6.  a;  =  —  5,  y  =  —  5,  z  =  —  5. 

4.  x.  =  —  2,y  =  3,  z  =  l.  7.  «  =  4,  x  =  5,  y  =  6,  z  =  7. 

5.  x  =  8,  y  =  -  3,  g  =  -  4.       8.  a;  =  3,  y  =  - 1,  s  =  0. 

9.  jc  =^  (b  +  g  —  a),  y  =  ^  (a  +  c  —  b),  y  =  ^  (a  +  &  —  c). 

10.  x~^,y  =  7-£,z=^     11.  a  =  -24,y=-48,g=60. 

12.  u  =  —  7,  x  =  3,  ?/  =  —  5,  z  =  1. 

_52+c2-ft2       _a2+c2-&2       _a2  +  &2-c2 

13,  *~       2bc    ~,V~       2ae      ,Z~~    2ab      ' 

14.  x  =  -,y  =  -^,z=--.  15.  a;  =  li  y  =  -X\,  «='l. 

16.  x  =  ab  c,  y=ab  +  ac  +  bc,  z  =  a  +  b  +  c. 

a  +  1  f- 

17.  x  =  7,  y  —  —  3,  g  =  —  5.     18.  as  = ,y—a  —  c,z-- 


c  a 

Art.  195 ;  pages  129  to  133. 

4.  A,  30 ;  B,  20.  5.  ^.  6.  Cows,  49 ;  oxen,  40. 

7.  A,  $140;  B,  $70.  8.  A,  98;  B,  15.  9.  32  and  18. 

10.  Man,  24 ;  wife,  18.  11.  Worked,  6 ;  absent,  4. 

12.  Horse,  $96;  chaise,  $112.  13.  A,  $96;  B,  $48. 


432  ALGEBRA. 

14.  16  days.  15.  13J  bushels  at  60  cts. ;  26|  at  90  cts. 
16.  Wheat,  9;  rye,  15.  17.  Income  tax,  $20;  assessed  tax,  $30. 
18.  A,  $500;  B,  $700.  19.  30  cents  ;  15  oranges. 
20.  1st,  8  cts. ;  2d,  7  cts. ;  3d,  4  cts.  21.  Better  horse,  $40  ; 
poorer,  $30;  harness,  $50.        22.  10,  22,  and  26.        23.  246. 

24.  A,  $2000;  B,  $3000;  C,  $4000;  D,  $5000. 
25.  A,  45;  B,  55.  26.  A,  $20;  B,  $30;  C,  $40. 

27.  Whole  sum,  $120;  eldest,  $40;  2d,  $30;  3d,  $24;  4th,  $26. 
28.  Length,  30  rods ;  width,  20  rods ;  area,  600  sq.  rods. 
29.   Going,  4  hours  ;  returning,  6  hours. 
30.  A,  9f  days  ;  B,  16  ;  C,  48.      31.  1st  rate,  6  p.c. ;  2d,  5  p.c. 
32.  15  miles  ;  5^  miles  an  hour.  33.  30  miles  an  hour. 

34.  A,  5 ;  B,  6.   35.  First,  22 ;  second,  10.         36.  A,  8 ;  B,  6. 

Art.  197;  pages  136  and  137. 

.  a  b  c  _     .     ,  _      m  a         ..     n  a 

4.  — — .  5.  li  hours.  6.  and . 

ab  +  ac  +  bc  vi-\-n  m,-\-n 

7.  12  and  8.      8.  -^-.      9.  12.     10.      10°"  11.  $2100. 

b  —  a  rt+  100 

12.  100>-i>),     13.  m.     14.  lst,g(c-6);2d;ft(a-r). 

p  r  ~  a  —  b  a  —  b 

,.-.».     ^,.,/n         , „    b  +  d  ^„    am+bji  +  cp 

15.  1st  kind,  5 ;  2d,  10.        16.  — — .        17.  -  -r-    — . 

a  —  c  a  +  b+c 

*  amt  .    -o  ant'  n  aPt" 


mt  +  nt'  -\-pt'n      '  mt  +  nt'+ptfn     '  mt+nt'+pf' 
Art.  205;  page  141. 


5 


3.-2  rods.      4.  -    ^ .      5.  105  and  —  15.      6.  In  —  30  years. 

7.  A,  .-$1500;  B, -$500;  that  is,  A  was  in  debt  $1500, 
and  B  $  500.  8.  Man,  $  3 ;  son,  —  $  0.50  ;  that  is,  the  man 
was  at  an  expense  of  50  cents  a  day  for  his  son's  subsistence. 


ANSWERS  TO  EXAMPLES.  433 

Art.  225;  page  152. 

4.  x  >  5.       5.  x  >  15,  x  <  20.      6.  4.      7.  x  >  6|,  y  >  2|. 

8.  a;  >  c,  x  <  d.  9.  *  >  9|,  7/  <  12£.  10.  19  or  20. 

11.  Any  no,,  integral  or  fractional,  between  8  and  15.      12.  60. 


Art.  229;   page  155. 

2  7  2 

1.  a3-3a2b  +  3ab2-b3.  2.~-2+—i. 

3.  1  +  3  a2+  3  b2+  3  a4+  6  a2  62  +  3  b*  +  a&+  3  a4  b2+  3  a2  64+  66. 

4.  a2  +  2am  —  2an  +  m2  —  2mn  +  n2. 

5.  a*m  _  4  a3m  +  »,  _|_  (5  a2m  +  2n  _  4  ((m  +  3a  +  a4n# 

6.  a5  +  5  a4  5  +  10  a3  b2  +  10  a2  bs  +  5  a  64  +  65. 

Art.  230 ;  page  156. 

3.  4a:4+12ar+25ar2+24z  +  16.     4.  4a:4-12a;3  +  lla;2-3a:  +  |:. 
6.  a;6+4a:5+6a;4+8a;3+9a:2+4a;+4.     7.  l-4a;+10a:2-12a;3+9a:4. 

8.  1  +  2  a;  +  3  ar  +  4  x3  +  3  a;4  +  2  a:5  +  a:6. 

9.  x6-8x5  +  12  x*  +  10  a;3  +  28  x2  +  12  x  +  9. 

10.  4  x«  +  4  a:5  +  29  x"  +  10  a;3  +  47  a:2  - 14  x  +  1. 

11.  a;6  +  10  x5  +  23  xi  -  6  a:3  +  21  x2  -  4  a;  +  4. 

12.  9  a;6- 12  x5  -  2  xi  +  28  x3  - 15  a;2-  8  a:  +  16. 

Art.  231;  page  157. 

2.  a^+Qa'b +  12  a2b2+U3.    3.  8m3+60?7r?z+150mw2+125»3. 

4.  27  a;3 -108  a;2 +144  a; -64.  5.  8  a;9  -  36  a;6  +  54  x3  -  27. 

6.  64 x6 -48  x5 y  +12  a;4?/2  —  a;3?/3. 

7.  27  a;3  ?/3  +  135  a  b2  x2  y2  +  225  a2  bAxy  +  125  a3  b6. 

Art.  232;  page  158. 

3.  a;6  -  3  x5  +  5  a;3-  3  x  -  1.  5.  8  -  24  a;  +  36  x2 -  32  a:3 
+  18  xi  -  6  a;5  +  a;6.                   6.  1  +  3  x  +  6  a;2  +  10  a;3  +  12  a;4 


434 


ALGEBRA. 


+  12  x5  +  10  x«  +  6  x1  +  3  x8  +  x\  7.  8  x»  - 12  xs  +  30  a;7 

-  61  x*  +  GG  x5  -  93  a4  +  98  a;3  -  63  a2  +  54  a;  -  27. 


Art.  239 ;  pages  162  and  163. 

2.  2  a;2  -  x  -  1.  5.  3  -  2  a;  +  a;2.  8.  3  ar  -  4  x  -  5. 

3.  2  a2  -  4  a  +  2.  6.  5  +  3  x  +  x\  9.  2  x2  -  5  a;  +  8. 
1 


4.  m  +  1 


m 


7.  1  —  7» -2 a;2.      10.  a  -b 


c. 


U.  x-2y  +  3z. 

13.    a-+  S fr—, :  4 


a; 


a;2      a;3 


i3 


8  a3      16  a5 
15.  a  +  s 


12-1  +  2-^+16 


2  a      8  a3 


+ 


x° 


16  a5 


Art. 

241;  page  166. 

2. 

523. 

7. 

95' 

12.  900.8. 

17. 

13.15295 

3. 

214. 

8. 

1.082. 

13.  .4125. 

18. 

.88192. 

4. 

327. 

9. 

21.12. 

14.  1.41421. 

19. 

.43301. 

5. 

5.76. 

10. 

.083. 

15.  2.23607. 

20. 

.57735. 

6 

.97. 

11. 

.00328. 

16.  5.56776. 

21. 

.53452. 

1.  3.3166. 

2.  1.732051. 


Art.  242;  page  168. 

3.  7.81024968.  5.  27.94638. 

4.  11.446.  6.  113.7234. 


Art.  243;  pages  170  and  171. 
2.1-2?/.       4.  4:x-3ab.  6.  y2  —  y-1.    8.  a;2-2a;  +  l. 


3.  2ar  +  3.      5.  a:2  +  2a;-4.  7.  x  +  -. 

x 


9.  a  +  b  +  c. 


10.  2  a;2 


3a;-l. 


115 

11.    X+'„    -o-K— R  + 


3  a;2      9  a;6  '  81a;8 


12. 


x 


ANSWERS   TO   EXAMPLES. 

435 

a? 

aG      5  a?                    g  0  2     1        1 

5 

3x2 

~9x5    Six8      '"             .        4z4    32z10 

768x16     "* 

Art.  245;  page  173. 

2.  123.  5.  «.  8.  1.442.  11.  .855. 

3.  .898.  6.  3.72.  9.  1.913.  12.  .420.* 

4.  11.4.  7.  .0803.  10.  5.963.  13.  .561. 

Art.  247;  pages  175  and  176. 
2.  to2  —  2m  —  4.     3.  a2-ax  +  x2.     4.  2x  — 1.     5.  x'2  —  x  +  1. 

Art.  248 ;  page  176. 
1.  2x-3y.  2.  cr-1.  3.  to2-2to-3. 

Art.  257 ;  pages  180  and  181. 

4.  c^.  5.  af*.  6.  *»*.  9.  -  6  a  J*K 

11.  a4  6~4  -  2  +  a-4  i4.  12.  a  -  b. 

13.  a-5-3a~3b2+  a-2bs-2a-1bi.    14.  18a2i2+10  +  2«-2&-2. 

15.  2ar12/-10zZ/-1  +  8.ry-3-         16.  2-4x"V  +  2affy. 
17.  6x2-7^-19af*+5a;+9^-2^.     18.  32aZr2-50+18a-1  &2 


Art.  258;  pages  182  and  183. 

5.  <T*.  6.  to5.  7.  x12\  8.  ^ ' 

11.  J  +  Jb^  +  Jb^  +  Jb^  +  bl      12.  a-^^^  +  i-2 
13.  x-3  f  -x~2y  +  2x-\  14.  a>* y-1  -  3  +  4 x~*  y, 

15.  x-1y-2-a;-27/-3-x-3y-4.       16.  2x^y~?J-x~^y-x~^ >*. 

Art.  260  ;  page  184. 
6.  x*.         7.  c~K        8.  to-1.      9.  2T3.       10.  a~3.       11.  w"1 


436  ALGEBRA. 


Art.  262;  page  186. 

'        1  n       12500 

3-9-  5-ioooo-       7-4'  9±-^~ 

4.  ±216.  6.  ±-^.        8.-243.  10.  ±£. 


Art.  263 ;  pages  186  and  187. 
5.  3  3-2 y  _  2  x-1  -  2T1-  6.  2  aj^  +  x  y~^  -4^  y~*. 


./•' 


7.  a?t"yT*-2-+aT*y*.       1L  2y*-y*ar1.         12. 
13.  a;-2"6.  14.  a*».  15.  a31.  16.  a-^.  17.  sc. 

01    •*  +  «*  oo    h^d'-afd*  9q    5 »"  (»»-!) 

21     1_3«3  *         ^    ab*#  +  aP*'  3a        ' 


Art.  267;  page  189. 
2.  ^27,  #16,  #25.  3.  $625,  ^216,  ^49. 

12.^ 12, 12 „      15,— —  15,-———-  15,. 


4.  V*V,  V^4^  VVs3-  5.  #32  a5,  #27  &3,  #64c3. 


12. 12, 


6.  #cr+2a6+62,  \/a3-3a-b+3ab2-b3.     7.  #a°-3a4.z2+3a2;c4-x6, 
fa8-2«3x3  +  x6.  8.  y^3.  9.  #2.  10.  #4. 


Art.  269  ;  page  190. 


5.  \l$ab\  6.  #a&2.  7.  v/(|f3)- 


Art.  270  ;  page  191. 


11.  3xyS/2xy'2-3xhj.  12.  (a  — 3)^a.  13.  (x  +  y)^x  —  y. 
14.  (2  x  +  3  a)  f  6a.  15.  4  a  &  #3a&2  +  5&.  18.  ^6- 
1    ._  „„    1   ,„.  „.    2a 


19.  ^i/30.  20.  Ji/21.  21.  ^f  y/3.  22.  r>#6*. 

6  b  y  J 


ANSWERS  TO  EXAMPLES.  437 


a\j  ab  c 


2  (a  +  x)  6-  (a  +  6) 

Art.  272;  page  192. 

7.  v^x       s-  ft^F-        9-  y/(^)- 

Art.  273;  pages  193  and  194. 


«/.  o   38 


3.  10 si 2.  4.  12 v/ 3.  5.  9^2.  6.  ^y/o 

7-  |\/6.  8.  |j2  +  ^18.       9.4^5.         10.  i^- 

11.6asJ3a.      12.  ^V3-  13.  ^2  +  ^3. 


14.  2  yV2  -  tf.  15.  (2  a  -  5  b)  \Jl  x. 

Art.  274;   pages  195  and  196. 
5.  \f^¥x-\  6.^4500^  7-  ^(g^ga)- 

8.  y/5^  .  10.  a;  +  v/z  -  6.  11.  21  x  -  38  v^  +  5. 

12.2.  13.-1.  U.  x-ij-z  +  2\jy^. 

15.4  +  2^10.  16.  56+12v/35.         17.36-32^15. 

18.  ax-x\  19.  m  +  TO.  20.  14  —  4^6. 

21.  147  +  30  v/24.     22.  1  +  2  a  %/l-a2.     2-3.  2  a-  2  \/a2-&2. 

Art.  275 ;  page  196. 

b/16 
V/243' 


e/8 

V9- 


7.  {/"  8.  {18.  9.  ^ 


438  ALGEBRA. 

Art.  276;   page  197. 
3.  ^125.        4.  y/7.         5.  2304  x\         6.  a4*2.        7.  \j~a~^-~b. 


8.  SlaHx  Sjbx.  9.  a:2  +  2x  +  l.  10.   16 a:2 - 48. 


Art.  277;  pages  198  and  199. 

12,- 


3.^2.         4.  si 2.         5.<)a  +  b.         6.  V*  -  1.  7.  y/2 

8.  ^3.  9.  v/3.  10.  ^^V-  n-  V2- 


Art.  278 ;  page  200. 

.    3s/2  .    ^4V2  x    5\/2  a    2°\/3tf 

O.    pr — .  4.    — s .  0.   — ^ — .  D.         5 

2  2a  2  3a 


Art.  279;  page  201. 

3.   12~4y/2  .  4.  5  +  2  V  3.  5.  2  ^  6  -  5. 

a  +  2_^b'+b  16  +  7  y/ 10 

6'  "        «-fi  ^  ~*  7*  "  13 


_    a  —  2  ^ ax  +  x  0        a  +  3  +  3  \la  +  1 

a  —  x  a 


a+^a*-x2 


10.  2a2-l-2aV/«2-l-       H.  •     12.  y^-l-ff2. 

a; 


1Q    x*-2  +  x^x3-4:  14  a:  -  24- 11  Vs8- 2a; 

2  18  —  o  x 


Art.  281;  page  202. 
2.  .894.  3.  7.243.  4.  3.365.  5.  .101. 

Art.  286;   page  204. 


4.  —  8  v/6.  5.  12\/ab.  6.  46.  7.  2. 

8.  -abc  V^l.         9.  a2  +  b.  10.  12.  12.  sJB. 


ANSWERS  TO  EXAMPLES.  439 

13.  y/2.  .     14.  y/5.        15.  ^3.        16.  \f^l.        17.  1  +  <f^2. 

18.  2(f-b)  .        19.  1  _  4  y/^3.  20.  - 100  -  18  tf=2. 

or  +  b  T 

Art.  293;  pages  207  and  208. 


5.  v/T  +  y/5.        8.  5  +  y/lO.     11.^15-^5.14.3-2^-2. 


6.  v/21-v/3.      9.  3-v/3.       12.3  +  ^5.       15.  5  +  3  \/^2. 


7.  3  +  v/7.         10.  v^5-v/3.    13.  7-3  y/ 2.    16.  6-V-l. 


17.  ^m  +  n—  \jm  —  n.  18.  x  —  SJax.  19.  3  +  v/2. 

20.  v/2-1.  21.  2-^3. 

Art.  297;  pages  209  and  210. 


4.  17. 

9.  4.            14.  4.            19.  -1.        24.  4. 

5.  19. 

10.  5.           15.  81.         20.  -3.       25.  5. 

6.  7f. 

11.  -2.        16.  4.           21.  4.            26.  3. 

7.  2. 

12.  |.           17.  8.           22.  12.         27.  6. 

8.  4. 

13.  4.           18.  -3.       23.  25.         28.  39. 

29. 

3£.           30.  3.            31.  6.            32.  3«-l 

Art.  303 ;  pages  212  and  213. 

2.  ±3. 

4.  ±y/ (_?).   6.  ±7.           8.  ±1.      10.  ± 

3.  ±5. 

5.  ±1.                 7.  iy/11.      9.  ±1.     11.  ± 

Ifr  —  IA 

12.  ±^\L-±y                   13.  ±v/a  +  ft. 

Art.  310 ;  pages  220  to  222. 
10.  5  or  -  7.  11.  11  or  -  2.  12.  5  or  3. 


440  ALGEBRA. 

13.  -  5  or  -  13.  29.  -  4  or  -  1.  45.  2. 

14.  ^or-|.  30.  2  or  i.  46.  4  or  0. 

15.  2  or  -.  '  31.  4  or  -  If.  47.  3  or  -  2. 

16,_l0r-|.  32.  4±2y/3.  48.  -2ov~ 
do  bo 

17.  ^^^  33.3or-l.  49.  ±2 


12  --■  —      ~  .y/8' 

18.  17±y337.  34.  2  or  - 1 .  50.  25  or  3. 

4  7 

19.  --  or  -1  35.  7  or  | .  51.  6  or  -2. 

o  2  6 

20.  lor-7,  36.  4or-^.  52.  -or--. 

4                                 4  a.  c 

21.  =f  or  -  2.  37.  - 10  ±  v^78.  53.  a  ±  b. 

o 

0Q    1±V409  3  a        a 

**.  £ .  38.  —  3i  or  —  2^.  54.  — — -  or  -  . 

b  ""42 

23.  —  or  -  .  39.  1  or  —  .  55.  —  a  or  —  b. 

4         Z  oh 

24.  3£  or  - 1.  40.  1  or  £  .  56.  11  or  18£. 

25.  13  or  -  2.  41.  5  or  ^ .  57.  5  or  -  3. 

5 


26.  I  or  i .  42.  18  or  3.  58.  12±J        . 
2       14  5 

27.  1  or  3i.  43.  —  2  or  —  .  59.  a  —  ft  or  —  a  —  c. 

28.  -  4  or  -  ** .       44.  -  3  or  2*.  60.  ^=*  or-3^. 

_,     a  +  b  a  —  b 

bl. or  — — r  . 

a  —  o  a  +  0 


ANSWERS   TO  EXAMPLES.  441 

Art.  311 ;  pages  224  to  227. 

4.  12  rds.     5.  40000  sq.  rds.,  and  14400  sq.  rds.     6.  9  and  6. 

7.  16  and  10.  8.  16.  9.  3  inches.  10.  $  30.  11.  14  and  5. 
12.  $2000.  13.  18bbls.,  at  $4  each.  14.  256  sq.  yds.  15.5. 
16.  7  and  8.     17.  7,  8,  and  9.      18.  Length,  125 ;  breadth,  50. 

19.  9.  20.  3712.  21.  80.  22.  20. 

23.  Area  of  court,  529  square  yards ;  width  of  walk,  4  yards. 

24.  36  bu.  at  $1.40.  25.  Larger,  $77.17^;  smaller,  $56.70. 
26.  1st,  14400;  2d,  625 ;  or,  1st,  8464 ;  2d,  6561.  27.  84. 
28.  6.                        29.  Larger  pipe,  5  hours  ;  smaller,  7  hours. 

30.  38  or  266  miles.  31.  70  miles. 

Art.  314 ;  pages  230  to  232. 

5.  ±3or±V-13.  6.  ±lor±J_  7.  lor -2. 

I  y  5 

8.  ±  1  or  ±  -  .  9.  ±  7  or  ±  5.  10.  ^3  or  -  ^23. 

11.  ±  8  or  ±  d(-  ^ .  12.  4  or  ^49.  13.  4  or  1. 

14.  243  or  -  J'  (285).  15.  4  or  1\.  16.  49  or  25. 

18.  2,  -  2,  3,  or  7.  19.  3  or  -  1.  20.  ±  1  or  ±  2. 

21.  2  or  -  3.         23.  1,  -  1,  5,  or  7.  24.  2,  -  3,  4,  or  -  5. 

25.  1,  2,  -  5,  or  8.  26.  1,  - 1,  -  6,  or  -  8. 


28.  3,-|or      3±4V/     55.  29.  8,  -2,  or  3  ±  y/ 110. 

80.?,-?,  or"3±22^3.  31.  1,9,  or  5±2N/2. 

32.  0,-5,  1,  or-^. 


Art.  317;  page  234. 
2.  x  =  2}y—±l;  or,  x  =  —  2,  y=±l.     3.  x  =  4,  y  =  ±  5 ; 


442  ALGEBRA. 

„      .  1  1  11 

or,  x=  — 4,  y=±5.     4.  «  =  g,  y  —  ±^  °r'  a;=—  3'  ^=±2' 

1  1 

5.  x  =  3,y  =  ±p  or,x  =  —  3,y  =  ±g. 


Art.  318;  page  235. 

2.  a?  =  7,  y  =  —  8;  or,  x  =  —  8,  y  =  7. 

3.  a;  =  5,  y  =  —  2  ;  or,  a;  =  —  2,  y  =  5. 

4.  x  =  3,  y  =  4 ;  or,  cc  =  —  4,  y  =  —  3. 

5  5 

5.  x  =  S,  y  =  2>  or>  cc=:~2'  y:=  —  8* 

1  5 

6.  x  =  2,  y  =  4 ;  or,x  =  —  ^,y  =  ^. 

7.  ar  =  2,  y  =  —  3 ;  or,  x  =  3,  y  =  —  2. 

8.  x  =  1,  y  =  2  ;  or,  a;  =  2,  y  =  1. 

Q  Q  9  15  62 

9.  a?  =  3,  y  —  2;  or,x  =  —  —,y  =  —. 

10.  a?  =  9,  y  =  6 ;  or,  x  =  —  6,  y  =  —  9. 

11.  a;  =  2,  y  =  9 ;  or,  x  =  9,  y  =  2. 

12.  a;  =  9,  y  =  3 ;  or,  x  =  —  3,  y  =  —  9. 

13.  x  =  6,  y  =  —  4  ;  or,  x  =  —  4,  ?/  =  6. 

14.  a;  =  3,  y  =  2;  or,  a;  =  —,  y  =  —  —  . 

15.  x  =  5,  y  =  3  ;  or,  a;  =  —  3,  y  =  —  5. 

16.  x  =  3,  y  =  —  7 ;  or,  a;  =  —  7,  y  =  3. 

Art.  319;  page  238. 

4.  x  =  3,  y  =  4;  a;  =  4,  y  =  3;  x  =  — 3,  y  =  — 4;  or,  x  =  —  4, 
y  =  — 3. 

U.  &  =  6,  y  =  7 ;  a;  =  7,  y  =  6 ;  a;  =  —  6,  y  =  —  7;  or,  a;  =  —  7, 
y  =  —  6. 

6.  x  =  2,  y  =  —  3 ;  or,  x  =  —  3,  y  =  2. 


ANSWERS   TO   EXAMPLES.  443 

7.  x  =  —  1,  y  =  4 ;  or,  as  =  —  4,  y  ==  1. 

8.  #  =1  3,  y  =  —  2  ;  or,  x  =  —  2,  y  =  3. 

9.  a;  =  4,  y  =  —  7  ;  or,  a;  =  7,  y  =  —  4. 

10.  x  =  5,  y  =  6 ;  or,  cc  —  6,  y  =  5. 

11.  x  =  5,  y  =  2 ;  or,  a;  =  —  2,  y  =  —  5. 

Art.  320 ;  pages  239  and  240. 


2.  x  =  2,  y  =  2>  x  =  -2,!/  =  -2>  x  =  \  5>y  =  —  2\ 


2 
5; 


5  g 

3.  x  =  2,y  =  3;  x  =  —  2,y  =  —  S-  x  =  ^^r,y  = 


^31  'y~     v'31' 
5  6 

y/31'  y"~^31.' 

4.  aj  =  3,  y  =  l;  a>=  —  3,  y  =  —  1;  x  =  2^2,  y  =  \/2; 

or,  ^  =  -2^2,  y  =  —  ^2. 

5.  £  =  3,  y  =  5;a;  =  —  3,  y  =  —  5;  x  =  -,  y  =  y5  or,  z  =  —  -, 

13 

6.  a;  =  2,  y  =  —  l\  x  =  —  2,y  =  l-,  x  =  —-r ,  y 


5  7 

7.  a;  =  2,  y  —  1 ;  a;  =  —  2,  ?/  =  —  1 ;  x  =  7,  y  =  — 19 ;  or, 

x  —  —  7,  y  =  19. 

Art  321;  pages  243  and  244. 

5.  x  =  l,  y  =  8;  or,  x  =  8,  y  =  1. 

6.  a;  =  4,  y  =  9  ;  or,  x  =  9,  y  =  4. 

7.  a:  =  2,  ?/  =  3 ;  or,  a:  =  3,  y  =  2. 


444  ALGEBRA. 

8.  x=3,  y=4  ;  3=4,  y=3 ;  x=-±  +  \J^11,  y=-A-\^H  ■ 

or,  x  =  —  4  —  \/—  11,  y  =  —  4  +  y^— 11. 

9.  a:=4,y=5;a:=16,  y=-7  ;  3=- 12+^/58,  y=-l— ^58 ; 

or,  x  =-  12  -  y/58,  y  =  -  1  +  y/58. 

10.  a;  =  4,  y  =  2  ;  cc  —  —  2,  y  =  —  4  ;  or,  a;  =  0,  y  =  0. 

11  o  a  605  20  io  -,  3 

11.  x  =  9,  y  =  4;  or,  *=.— ,  y  =  — .  12.  a- =  1,  y  =  -. 

13.  x  =  3,  y  =  2 ;  or,  a;  =  2,  y  =  3.  14.  x  —  9,  y  =  4. 

15.  3  =  1,  y=-3;  a=-3,  y=l;  a-=l+V^^2,  y=l- V^2; 

or,  a;  =  1  —  \/-2,  y  =  1  +  ^~2. 

1«         1  o         o  -,  3+V-55        -3+V-55. 

16.  x=l,y=—2;x=2,y=—l;x= 1 ,y= j ; 

or,x  = ,y  = y. . 

17         9         o  o  o         -1+3V^3        1+3^. 

17.  x=2,y  =  3;a:=-3,y=-2;a:  = ^ ,y= J , 

-1-3  y/^3           1-3  V/'=r3 
or,  a;= ^ »  ?  = ^ ■ 

18.  ,=3;y=2;,  =  2,,=3i,=^±^,,==^.9; 

_-9-v/309  -9  +  ^309 


12      ",y~  12 

19.  3=1,  y=-3;  3=-l,  y=3;  3=141,  y=3f;  or,  3=-14f, 

20.  3  =  2,y  =  3;  or,3  =  2f>y  =  lf. 

01,100  4  22  59 

21.  a:  =  4,  y  =  2,  «  =  3;  or,  x  =  -,ij  =  —  ,  z  =  -g-- 

22.  a;  =  1,  y  =  2,  z  =  4  ;  a;  =  —  1,  y  =  —  2,  s  =  —  4  ;  x  —  9, 

y  =  —  6,  s  =  4 ;  or,  x  =  —  9,  y  =  6,  z  =  —  4. 

Art.  322;  pages  246  to  248. 

4.   12  and  7,  or  —  12  and  —  7.      5.   11  and  7,  or  —  11  and  —  7. 

6.  A,  $2025;  B,  $900;  or,  A,  $900;  B,  $2025. 


ANSWERS   TO   EXAMPLES.  445 

7.  A,  25 ;  B,  30.  8.  Length,  150  yds. ;  breadth,  100  yds. 

9.  13  and  6.        10.  A,  $15;  B,  $80.        11.  10  lbs.,  at  8  cts. 
12.  A,  $  5 ;  B,  $  120.  13.  Duck,  $  0.75  ;  turkey,  $  1.25. 

14.  Price,  $  1600  ;  length,  1G0  rods ;  breadth,  40  rods. 
15.  Larger,  864  sq.  in. ;  smaller,  384.     16.  A,  $  275  ;  B,  $  225. 
17.  1st  rate,  7  p.c;  2d,  6.     18.  A,  40  acres  at  $  8  ;  B,  64,  at  $  5. 

19.  Distance  of  towns,  450  miles  ;    A,  30  miles  a  day  ;    B,  25. 

20.  3  and  1 ;  or,  2  +  sj  7  and  2  -  y/  7.  21.  Larger,  12  ft. ; 
smaller,  9.        22.  Width  of  street,  63  ft.;  length  of  ladder,  45. 

23.  B,  15  days ;  C,  18  days. 
24.  Length,  16  yds. ;  width,  2  yds. 

Art.  328 ;  page  253. 

3.  (a: +  60)  (a; +  13).  6.  (x  +  13)  (a? -3).      9.  (4a;-l)  (2a; +  5). 

4.  (x -9)  (x -2).     7.  (jc-5)(2a;  +  3).    10.  (x - 3)  (4 x - 3). 

5.  (a; -10)  (a; +  6).   8.  (7 a; +  3)  (3 a; +  7).    11.  (a; +  2)  (2  a; -3). 

12.  (3a:-2  +  v/3)(3a-^2-v/3).     13.  (y/17  +  4  +  a-)  (^17-4-3:). 
14.  (7a3  +  l  +  2v/5)(7a;  +  l-2v/5). 

Art.  329  ;  page  254. 

2.  ar  +  a;  =  2.  5.  3  x2  -  2  x  =  133.       8.  3  x1  +  17  x  =  0. 

3.  a;2 -9  a; =-20.     6.  21.x2  +  44 x  =  32.     9.  a:2-2a;  =  4. 


4.  5a;2-  12a; = 

9.      7.  6ar+35a;=-49.    10. 
Art.  330 ;  page  255. 

x2 — 2  m  x=n— m4 

7.  0ory. 

8.  0  or  —  4.        9.  0  or  ±  3. 

in       5      x 

lL-*orl 

a        c 

12.  ±  2  or  ±  3. 

1          5 

13--3°r±2 

14.  ±sja 

or — .       15.  0,  - 

5    7             1 
■2'3'°r~4- 

16.  2,3,-3,  -4,  I,  or  -5. 


446  ALGEBRA. 

Art.  331;  page  256. 


3.  (x  +  y/2  x  +  1)  (x  -  \/2  x  +  1). 

4.  (a;  +  \Jx  +  1)  (x  —  \J  x  +  1). 

5.  (a  +  ^5lTb  +  b)  (a  -  \j~h~ab  +  b). 

6.  (x°~  +  3  x y  +  f)  (x2-3xy  +  f). 

7.  (x  +  1  +  S/Sx  +  2)  (x  +  1  -  \J~3x~+2). 

8.  (m2  +  mn  +  w2)  (m2  —  m n  +  to2). 


Art.  332 ;  page  256. 


2. 


7.  ^  7  *  V71- 1  ™  -v/7±V/:-l 


or 


2^2  2^2 

Art.  357;  pages  269  and  270. 

1.  4.        2.  11.         3.  £.         4.  1  J.         5.  ±4.         6.  ±  12. 

7.  ±  14.  8.  25  and  20.  9.  23  and  27.  10.  12  and  15. 
11.  8  and  18.  12.  26  and  14.  13.  17  and  12.  14.  12  and  8. 
15.  First,  1:2;  second,  2  : 1.  16.  Females  :  males  =  4:5. 

17.  8  :  7. 

Art.  365;  page  273. 

2.4.         3.  „  =  8*  4*  5.4.         6.  y=     14.    . 

°>  4  — 5cc 

7.  10  inches.  8.  3  (y/  2  —  1)  inches.  9.143. 


ANSWERS   TO   EXAMPLES.  447 


Art.  370;  page  276. 

3.  1=71,  £=540.     4.  Z=-69,  £=-620.  5.  1=57,  £=552. 

OQ  A9  3 

6.  l=-U5,  £=-2175.     7.  Z=^|,£=y.  8.  l=--,S=0. 

9.  l=-^,S=±.     10.  l=~,S=~.  U.l=5,S=17. 

4  2 


Art.  371;  pages  278  and  279. 

q  95  1 

4.  a  =  3,£  =  741.      5.  a  =  ^,i==--^-.       6.  <Z=-,£=39. 

7.  d  =  — i,  «  =  -?.    8.  a=o,d=-3.    9.  w=18,  £=411. 
12  4 

10.  «Z  =  — 8,  n  =  ll.     11.  w  =  30,  Z  =  80.     12.  ^  =  52,  a  =  4; 
or,  »  =  43,  a  —  —  5.         13.  n  =  l%l  =  -  43. 


Art.  372;  page  279. 

7    8        10    11  5        3        1  1 

A3'3"I'"3""  J,2'w,2'    '2'    '      2' 

4       2      3  _4  _5     5   Jl   -^    _^    _!?    -^    -?? 

*■      ^ — ^     4>     °-  7  '      7'      7'       7'       7'       7 

2    6    14    22  „    am  +  5    a(m-l)  +  2b 


5'5'  5  '   5  '  '    m  +  1  '  m  +  1 

Art.  373;  page  281. 
3.  2500.  4.  Last  payment,  $  103  ;  amount,  $  2704. 

5.  4.  6.  After  9  days,  at  a  distance  of  90  leagues. 

7.  4,  11,  18,  and  25.  8.  3.  9.  0.  10.  20  miles. 

11.  2,  6,  10,  and  14 ;  or,  -  2,  -  6,  -  10,  and  - 14.  12.  8. 


448  ALGEBRA. 

Art.  378;  page  284. 

4.  1  =  2048,  £=4095.  9.  1  =  -  —  ,  s  =  -^^ 

64  '  192 

64  2059  _  1  511 

5-  l  =  m>  S=  243-.  10.  1  =  ^,  S=~. 

6.  Z  =  2048,  £=1638.  11.  l^-~,  £=^. 

7    l--- L    s-3^  12    Z-        1       <?-      341 

*~     256' ^-256'  "'  ^-~768'^-~256- 

1  2047 

8-  l  =  zm>S=mi-  13^  =  192,5=129. 


Art.  379;  page  286. 

-           1    c         341  2     _        2 

4.  a  =  ^,  £  = 7r-.  5.  a  =  7.,    Z  = 


2'  2  ""  "-_3'  6561' 

6.  r  =  3,  £=2186;  or,  r  =  -3,  £=1094. 

1  2457 

7.  ^-j, /S=m.  8.  »  =  5,  £=121. 

9.  n  =  l,  r  =  \.     10.  w  =  6,  Z  =  -?S       1L  w  =  8,  a  =  -l. 

^  2 


Art.  380 ;   pages  287  and  288. 


3.  4. 

5   3 

160 
y-  19" 

4  ? 

3 

ft   15 

Art.  381 

8'i' 

;  page  288. 

,0.-12. 

3.|.   4. 

13 
27" 

5  n 
5-  15- 

ft   8G      7 

6-  165*   7' 

17       237 
150*     1100 

ANSWERS   TO   EXAMPLES.  449 


Art.  382;  pages  289  and  290. 

48  16   32    64  392781243  39 

'3'  9'  27'  81'  243"         2'  2'    2  '    2  '    2   '  °r'       2'  2' 

-|,  |,  -y.    5.  - 6,  -18,  - 54,  - 162,  - 486,  -  1458. 

927        81243  33333       3        3 

~4'  16'  ""64'  256*  4'  8'  16'  32'  64'  128'  256' 

3    3        _3_    _3^         3_    _3_        _3_ 
°r'~4'  8'  ~16'  32'       64'  128'  "    256' 


Art.  383;  page  291. 

3.  $  64.       4.  $  295.23.       5.  3100  ft.       6.  5,  10,  20,  and  40 ; 
or,  -  15,  30,  -  60,  and  120.  7.  -  4.  8.  TV 

Art.  386 ;  page  292. 

Z.E.         3.  -i.         4.      3  "  •» 


31'  78'  '       4'  '  arc-&?i  +  2&-a' 


2. 


Art.  387;  page  293. 
48     24    16    12     48      8      48 


125'  65'  45'  35'  145'  25'  155' 


5_5_5  ■         21     _7        21        21 

3'       4'"   3'"    2"  '"       '"  5  '      3'      13'      17" 

_     (m  +  1)  a  6        (m  +  1)  a  b  (m  +  1)  ab 


m 


b  +  a      '   m6  +  2a-6'  mH3«-2i' 


Art.  397;  pages  297  and  298. 
4.  Of  4  letters,  360  ;  of  3,  120  ;  of  6,  720  ;  in  all,  1956. 
5.  1680.   6.  3838380.   7.  358800.   8.  15120.   9.  120. 
10.  35.   11.  15504.   12.  31824.   13.  77520.   14.  648. 


450  ALGEBRA. 

Art.  403 ;   page  302. 

5.  1  +  5  c  +  10  c2  +  10  c3  +  5  c4  +  c5. 

6.  a6  +  6  a5  a8  +  15  a4  x6  +  20  a3  x9  +  15  a?xu  +Qa  xlh  +  xls. 

7.  xs-8x6y  +  24:Xi7f-S2x2f+16y\ 

8.  a7^-7a6i6c^  +  21a5i5cV2-35aH4c8^3+35aH8c4^4 

-21a2&2c5d5+7a&c6a,6-c7cf. 

9.  m12  +  18  m10  w2  +  135  m8  w4  +  540  m6  w6  +  1215  m4  w8 

+  1458  m-  n10  +  729  w12. 

10.  a- 10-20 a- 8a^+160 ar6x- 640a~  ^+1280 or 2ar-1024a^. 

11.  c™  +  8  c*'  cfl  +  28  c4  e^  +  56  c$~  <fi  +  70  c%  d3  +  56  c2  d^' 

+  28  c*  <#  +  8  J  eft  +  d6. 

12.  m~^  +  14»"^  w3+ 84 m-8»6+280  m-^"  w9+ 560m~*»12 

+  672  m"^  w15  +  448  j»~*  w"  4-  128  w21. 

13.  a-4-4ffl-3i2x^  +  6  a~2 Z>4 a;^  -  4  or1  66  a;  +  bs  sA 

Art.  404;  page  303. 

2.  5005  a6  a;9.  4.  - 19448  c10  d7.  6.  42240  x~3  yK 

3.2002  m6.  5.495  a8.  7.  262440  a2  ar7. 

Art.  405  ;  page  304. 

2.  1  -  4  x  +  2  a;2  +  8  x3  -  5  xi  -  8  xh  +  2  x6  +  4  x1  +  x\ 

3.  a;6  +  9  x5  +  30  a;4  +  45  x3  +  30  x2  +  9  x  +  1. 

4.  l-6a;  +  6a;2+16a:8  — 12a4-24a;8-8a:6. 

5.  l+5a;  +  5a;2-10a;3-15a;4+lla;5+15x6-10a;7-5x8+5x9-cc10. 


Art.  414;   page  309. 

3.  l-2a:  +  2a;2-2a:3  +  2a;4 

4.  3  +  19  x  +  95  x1  +  475  xa  +  2375  a;4 

5.  2-a;  +  3a:2-a:8  +  3a:4 


ANSWERS   TO  EXAMPLES.  451 

6.  l-2a;  +  2a;3  —  2x4  +  2x6 

7.  l-2x  +  ox2-16x3  +  4:7xi 

1      5x      7_x*      lTjc3      31_x4 

9.  2- 7  a;  +  28a;2  -  91  a;3  +  322  a;4 

2  a;      7  a;2      13  a;3      8  a;4 

io.  i  +  — -—      27  +  81  

1      3jc      a2      15k8     49a4 
1L  2  +  ~T  + IT*    16    +    32    


Art.  415;  page  310. 

2  a;-2      4  a;-1      _8_      16  a;      32  a;2 
2-  _3_+_9_+27+   81  +  243 

3,  £C-1  +  3  +  2x-5a;2-16a;3 

4.  x~2  —  x-1  -2 x  +  2 a;2 -4 a;3 


Art.  416;  page  311. 

a;      a;2      a;3      5  a;4  a;      3  a;2      3  Xs      3  a;4 

3*  1  +  2~8~+16~128""  ,1+2+    8      "16+128"' 

a;2      a;3      5  a;4  _ a _  s2      5  a;3      10  a;4 

3.  l_aj_— — - — g-...  «>.  i      3      9  -~81     -  243  •'• 

x4  „     .     a;     2  a:2    13  a;3      8  a;4 

4.  l_a;  +  a;2  +  a;3+ y...  7.  1  +  -+— — ^  +-^3  ... 


Art.  418 ;  page  314. 

0        3  2.41  _J_ 6_ 

2-  x~T2+x~=2'         x-2     x  'x-7      x-6' 

,    3         2  ,1  J_       7    _2 3_ 

«~T+8"  &'^T4  +  a;  +  l'     '"2»-5      Sas  +  l" 

Q        1  2  _J 1  1         4 

8"3+T£  +  3=!T        y"6(x  +  l)      2(a;-l)i"3(a;-2)' 


452  ALGEBRA. 


Art.  419 ;  page  316. 

o       11  1  „        1  4  4 

2-    — ^-  +  7— ^Tv2  +  7— "TV,-      4.-—  +  - —  + 


x  +  l      (x  +  l)2      (x  +  l)3'  x-2      (x-2f  '  (a; -2) 

3      2  3  5      3  6 


B' 


x-5      (x-5)2'  x  +  l      (x  +  l)'2      (x  +  l)3' 

3  5 


6. 

7. 


2  (2  a;  -  5)      2  (2  x  -  5)2 ' 
2  4  3 


3  a; +  2      (3x  +  2)2      (3x  +  2)3 


Art.  420 ;  page  317. 
_    2         3  5  .515 

2. —  -, — ;  .  4. 


x      x+2      (x+2)2'  'x      x*      x  +  l      (x  +  l)2' 

„11  1  1  .123         4 

3.    -+— T  +  —0  +  7— 2K3-        5.- ,+ 


x      x— 1      x—2      (x—2)'2'        '  x      x2      xs      x  +  5* 


■i 


x-2     2x-3      (2x-3) 
„   5      1       2         5  4 

7. S  +  -T-— 


X         X2         X3        X  +  l         (x+l)2' 

Art.  422;   page  320. 

o  2,84  /i  y      3?/      19?/6      19  y' 

3.  x  =  y  —  ?/2  +  ?/8  —  y 4  . . .     4.  x  = sL  -j £ —  . . 

J      J       J       J  2      16  +  128       128 

5.  x  =  y  +  y3  +  2  y5  +  5  y7  . . . 

vy       ;  2       i-       3  4       ■" 

7   x_?/,  ^,2^    17  y'  y     2y2      y3       14y* 

^~y+3+l5-+315"-     8'  *-3+27~~243~2T87" 


Art.  425;   page  325. 

-      §      5    |      .15   1,5    a   .       5     _a 
4.  a2  +  -a2  x  +  -^-a2  x'  +  z—a    2  xz  —  — -a   ^4 
«s  8  lb  128 


ANSWERS   TO   EXAMPLES.  453 


5.  1  -  6  x  +  21  x2  -  56  x3  +  126  x* 


3         12   2      52     3     234    4 
b.  i  +  5x  +  25x  +125x  +G25x  

i      1    _i.         1    _|  1     _i  5      -z 

7.  aJ  —-^a    *x  —  -^a    'X'  —  zr^a    2x6  —  zr^a    J  x* 

£  o  lb  lwo 

„    ,      1         2    „      14,       35 
8.1--x  +  -x*--x,  +  mX> 

9.  ar3  +  3a-4x  +  6  a~5 x2  +  10  a~6  x3  +  15  «"7  x*  .... 
10.  c"^—  c-3  df  +  c~*  d2  -  c-6  d3  +  c"1*'  d* 


11.  x    z —2xb  y —  Xs y2 —  -Xs  y3  —  ^x3  y* 

o  o 

I    3      15     7  35     3    a     315     jjl    6 

14.  m  +  6  in6  n-  +  -jr-  mJ  n6  +  -=-  mr  n2  -\ — —  m  J  ?i   . . . 

2  2  o 

in    -i      -i^         i     ™   o      o     1760    „      „     12320 

13.  1  —  10  xy-1  +  80  a;2?/-2 q-b'jt'H 5—  a?4?/"4 

2  2  8 

15.  a4  +  12  a5  gT2  +  90  a6  y~4  +  540  a7  jr6  +  2835  as  y 


—  8 


Art.  426 ;  page  326. 

33  aT'^'x''  315  a8  44  x^'  y* 

"~2048   '      128"'  6561  ' 


663  x   "V 


4.  84  m6.        6.  -   "   y  ■  8.  210  w^c"8. 

8192 

9.  _?5?a-^aj-6.      10.  3§x-3«y-Viz~1£. 


Art.  427;  page  327. 

3.  3.14138.       5.  9.94987.       7.  2.03054. 

4.  2.08008.       6.  1.96101.       8.  2.97183. 


454 


ALGEBRA. 


2. 
3. 


1  +  x 


Art.  435;  pages  331  and  332. 

4 -11  x  „    2  +  5«  +  5a2 


1  —  x  —  x2 ' 
a 


b  +  ex 


4. 
5. 


1  —  5  x  +  6  x2 

1+x 
1  -  2  x  +  x2 ' 


6. 

7. 


(1  +  a)3       ' 
3  —  x  —  6  x2 

l-2a;-a:2+2a;3 


3.  3. 

8.  225. 


8. 


l  +  2cc 
1  —  x  —  x2 


9. 


2  +  2  a;  -  3  x2 
l  —  x  +  x2  —  x& 


4. 


Art.  440;  page  336. 
14.  5.  30.  6.  1365. 


7.  5050. 


9. 


ni  +  2nz  +  n2 
11.  165. 


10. 


6  n6  +  15  m4  +  10  w8 


?« 


30 


12.  5525. 


3.  4.0514. 


1.  1.681241. 

2.  2.644438. 

3.  1.748188. 


Art.  443 ;  pages  338  and  339. 

4.  3.634241.  5.  2.23830.  6.  44.24. 

7.  $1,356. 


Art.  455 ;   page  344. 

4.  1.991226.      7.  2.225309. 

5.  1.924279.      8.  3.848558. 

6.  2.753582.      9.  2.702430. 


10.  3.489536. 

11.  4.191785. 

12.  4158543. 


1.  1.176091. 

2.  2.096910. 

3.  0.154902. 


Art.  456 ;   page  345. 

4.  2.243038.  7.  0.853872. 

5.  0.522879.  8.  1.066947. 

6.  1.045758.  9.  0.735954. 


Art  464;  page  350. 

2.  8.724276-10.     4.  9.470704-10.       6.  1.527511. 

3.  1.714330.  5.  0.011739.  7.  8  780210-10. 


ANSWERS   TO   EXAMPLES. 


4jj 


8.  4.812917.  11.  9.942550-10.    14.  4.89381. 

9.  7.013150-10.    12.  3  863506.  15.  1.718451. 

10.  2.960116.  13.  8  640409-10.     16.  7.4984240-10. 

17.  9.275374-10.  18.  1.9792784. 


2.  76. 

3.  .2954. 

4.  6.61005. 

5.  55606.5. 

6.  .011089. 


Art.  465  ;   page  352. 

7.  186  334.  12.  .034277. 


8.  .223905. 

9.  1000.06. 
10.  9.77667. 

•    11.  .00130514. 
17.  .00548803. 


18. 


13.  46.7929. 

14.  11.327. 

15.  8.63076. 

16.  .2070207. 
734.9114. 


Art.  466;   pages  353  and  354. 

1.  2.125240.  4.  3  108462.  7.  9.613158  -  10. 

2.  8.223962-10.      5.  9.594161-10.      8.  9.970036-10. 

3.  9.852169-10.      6.  7.315321-10.      9.  9.905232-10. 


Art.  468  ;   pages  356 

1  to  358. 

1. 

.0341657. 

13.   1.70869. 

25.  .580799. 

2. 

.650573. 

14.  .788547. 

26.  -.631188. 

3. 

13560.2. 

15.  .680192. 

27.  83.5656. 

4. 

.136085. 

16.  2.24328. 

28.  .297812. 

5. 

1.14720. 

•    17.  .296850. 

29.  98.4295. 

6. 

1.41421. 

18.  -.191680. 

30.  1.65900. 

7. 

1.49535. 

19.  .644849. 

31.  3  07616. 

8. 

.0655264. 

20.  .501126. 

32.  .867674. 

9. 

-1.97221. 

21.  1.09872. 

33.  -2.09389 

10. 

458.623. 

22.  1.06178. 

34.  46809.2. 

11. 

-.000113607.     23.  1.09328. 

35.  .588142. 

12. 

5.88336. 

24.  1.65601. 

36.  1.80446. 

37. 

.00323011. 

38.  .0334343. 

456 


ALGEBRA. 


The  following  are  the  values  of  the  expressions  in  Art.  468, 
when  calculated  by  seven-figure  logarithms  : 


1. 

.034165G8. 

13. 

1.708689. 

25. 

.58079S7. 

2. 

.0505727. 

14. 

.7885469. 

26. 

-  .6311888 

3. 

13560  27. 

15. 

.6801947. 

27. 

83.56558. 

4. 

.1360851. 

16. 

2  243284. 

28. 

.2978123. 

5. 

1.147203. 

17. 

.2968501. 

29. 

98.42-J91. 

6. 

1.414214. 

18. 

-  .1916795. 

30. 

1.658989. 

7. 

1  495349. 

19. 

.6443490. 

31. 

3.076162. 

8. 

.06552632. 

20. 

.5011282. 

32. 

.8676754. 

9. 

- 1.972211. 

21. 

1.098718. 

33. 

-  2.093891. 

10. 

458.5759. 

22. 

1.061780. 

34. 

46808.95. 

11. 

-  .0001136063 

.  23. 

1.093280. 

35. 

.5881412. 

12. 

5  883366. 

24. 

1.656005. 

36. 

1.804459. 

37.  .003230121 

38. 

,03343431. 

Art.  469  ;  page  359 

3.  , 

458156. 

5. 

-  .494903. 

7.  - 

-2.70951. 

4.  . 

185339. 

6. 

-  .260231. 

8.  - 

- 10.2341. 

The  results  with  seven-figure  logarithms  are  as  follows  • 

3.  .4581568.  5.  -  .4949028.  7.  -  2.709513. 

4.  .1853394.  6.  -.2602272.  8.  -10.23414. 


Art.  479  ;   pages  368  and  369. 


1.  7. 

3.  -  6.                  5. 

7. 

7.  6. 

2.  6. 

4.4           e. 

5. 

8.  7. 

9.  1.56937. 

13.  11.725  yrs. 

17. 

3.96913. 

10.  2.44958. 

14.  $9756.59. 

18. 

7.18923. 

11.  2.00906. 

15.  7  per  cent. 

19. 

-  2.4578. 

12.  $5421.33. 

16.  9.392  yrs. 

20. 

- 1.07009 

ANSWERS   TO   EXAMPLES.  457 

The  results  of   the   last   12   examples,    using   seven-figure 
logarithms,  are  as  follows  : 

9.  1.569369.  13.  11.725  yrs.  17.  3.969124. 

10.  2.449576.  14.  $9756.59.  18.  7.18922. 

11.  2.009056.  15.  7  per  cent,  19.  -2.457802. 

12.  $5421.35.  16.  9.392  yrs.  20.  -1.070092. 

Art.  489;  page  373. 
2.  3  and  -  5.     3.  a  and  |  (- 1  ±  y/^3).     4.  2  and  2.     5.  ±  4. 

6.  X3_6a.2_6a,_3  =  0      7>  ?and    5       8.  ?and-l 

3  2  4  5 

Art.  490;  page  374. 

2.  z3  +  9x2  +  23£  +  15  =  0.        4.  6ai3-lla:2  +  6*-l  =  0. 

3.  a8 -19  a: -30  =  0.  5.  cc4-  5x'  +  4  =  0. 

6.  a;4  -  10  x*  +  35  x2  -  50  cc  +  24  =  0. 

7.  cc3  - 13  x2  +  56  x  -  80  =  0. 

8.  x4-6a;3+5x2  +  12x  =  0. 

9.  12  z4  +  55  r3  -  68  x2  -  185  x  +  150  =  0. 


Art.  494;  page  375. 


5 


1.  Sum,  0  ;  product,  -  6.  2.  Sum,  -  ;  product,  12. 

3.   2±2v/2. 

Art.  504;  page  382. 

2.  ?/3  +  24?/2  + 191?/ +  498  =  0.    3.  y*  -  6?/  -  if  +  55y  -  76  =  0. 


P 


■: 


Art.  505;  page  383. 


2.   y2-^  +  <7  =  0.  4.  ^-15y+26  =  0. 

y      3  +  "2y==0,  5*   2/4-6y2-137/-9  =  0. 


458  ALGEBRA. 

Art.  513;   page  388. 

2.  1,  1,  and  6.  4.-1,-1,-1,  and  3. 

3.  2,  2,  and  3.  5.   2,  2,  2,  and  -  6. 

Art.  517;  page  390. 
2.    - 1,  1,  and  5.  3.    3.  4.    1.  5.   2. 

Art.  520;  page  392. 
3.  1  +  s/U.     4.  1  +  V15-      5-  -  (1  +  V^)-       6.   -  (1  +  $5). 

Art.  527;  page  399. 

3.  Three ;  respectively  between  0  and  1, 1  and  2,  and  —  1  and  —  2. 

4.  Three ;  two  between  1  and  2,  and  one  between  —  3  and  —  4. 

5.  One ;  between  2  and  3. 

6.  Four ;  respectively  between  0  and  1,  1  and  2,  2  and  3,  and 
—  2  and  —  3. 

7.  None. 

8.  Two  ;  respectively  between  2  and  3,  and  3  and  4. 

Art.  532;  page  403. 

3.  —  1,  - 2,  and  - 3.  9.  A,  and  1  ±  }f—i. 

4.  2,  -2,  and -3. 

5.  2,  4,  and  - 1  ±  y/^3. 

6.  - ,  4,  and  —  ^  • 

7.  2,  and  — — *-  . 

8.  3.  6,  and  -  2. 


10. 

1, 

2,  and  3. 

11. 

3 

9 

,  and  ±  2 

v/- 

-2. 

12. 

o 

13. 

3. 

14. 

3, 

4.  —  3,  and  ■ 

—  5. 

ANSWERS  TO  EXAMPLES.  459 

Art.  538;  pages  407  and  408. 
2        1   9±x/77        3  ±  y/5  l_^±V/^_2j?-3 

3.  -1,1,1,  or  =^?.         6.  2,i,-3,or-|. 

4.  ±lJ±V/=T,or^f:  3.    7.  l,5,i    or2±^3. 

y/33- 5  ±^42- 10  y/33  - y/33 - 5  ±  y/42  +  10  y/33 

8-  4  '  °r  ~I~ 

q    _i   1  +  y/5±V/2V/5jri~Q"    or    1-^5+^-2^5-10 
a>       *'  4  4 

-l-^5±y^5-10          -l+v/5±y/-2v/5-10 
iu.  w, 2~  ?  01    o 

Art.  541;  page  410. 

3.  -3or^±£3.  7.   lorl±YE5. 

2  2_ 

4.  4  or  1  ±  4  \/=3.  8.   3  or  1  ±  ^~  3' 

5.  3,  3,  or -2.  9.    2,  2,  or— 1. 

6.  1,1,  or  -11.  10.   $±-1(2. 

Art.  550;  page  417. 
2.2.09455.       3.7.61728.       4.  1.3569,  1.6920,  and -3.0489. 
5.  14.95407.         6.  2.2674  and  36796. 

7.  2.85808,-60602,  .44328,  and  -  3.90738. 

Art,  551;  page  419. 

2.  3.864854.  4.   2.4257.  6.   10.2609. 

3.  4.11799.  5.    .66437.  7.   8.414455. 

Art.  552;  page  420. 
2.    153209.  3.    1.02804. 


T  A  B  L  E 


CONTAINING  THE 


LOGARITHMS   OF   NUMBERS 


FROM   1   TO   10,000. 


No. 

Log. 

No. 

Log. 

No. 

Lo£. 

No.    Log. 

No. 

Loft. 

1 

0.000000 

21 

1.322219 

41 

1.612784 

61 

1.785330 

81 

1.908485 

2 

0.30103C 

22 

1.342423 

42 

1.623249 

62 

1.792392 

82 

1.913814 

3 

0.477121 

23 

1.361728 

43 

1.633468 

63 

1.799341 

83 

1.919078 

4 

0. CO 20 GO 

24 

1.380211 

44 

1.643453 

64 

1.806180 

84 

1.924279 

6 

0.698970 

25 

1.397940 

45 

1.653213 

65 

1.812913 

85 

1.929419 

6 

0.778151 

26 

1.414973 

46 

1.662758 

66 

1.819544 

86 

1.934498 

7 

0.845098 

27 

1.431364 

47 

1.672098 

67 

1.826075 

87 

1.939519 

8 

0.903090 

28 

1.447158 

48 

1.681241 

68  i  1.832509 

88 

1.944483 

9 

0.954243 

29 

1.462398 

49 

1.690196 

69 

1.838849 

89 

1.949390 

10 

1.000000 

30 

1.477121 

50 

1.698970 

70 

1.845098 

90 

1.954243 

11 

1.041393 

31 

1.491362 

51 

1.707570 

71 

1.851258 

91 

1.959041 

12 

1.079181 

32 

1.505150 

62 

1.716003 

72 

1.857332 

92 

1.963788 

13 

1.113943 

33 

1.518514 

53 

1.724276 

73 

1.863323 

93 

1 .968483 

14 

1.146128 

34 

1.531479 

54 

1.732394 

74 

1.869232 

94 

1.973128 

15 

1.17C091 

35 

1.544068 

55 

1.740363 

75 

76 

1.875061 

95 

1.977724 

16 

1.204120 

36 

1.556303 

56 

1.748188 

1.880814 

96 

1.982271 

17 

1.230449 

37 

1.568202 

57 

1.755875 

77 

1.88641)1 

97 

1.986772 

18  1.255273 

38 

1.579784 

58 

1.763428 

78 

1.892095 

98 

1.991226 

19  I  1.278754 

39 

1.591065 

59 

1.770852 

79 

1.897627 

99 

1.995635 

20  |  1.301030 

40 

1 .602060 

60 

1.778151 

80  1.903090 

too 

2.000000 

LOGARITHMS 


I   2   |   3   | 


8   |   9 


N.|   0 


I  1 


D. 


100,001)000 


4321 
8000 

012837 
7033 

021189 
6300 
9384 

0334  21 
7420 


000434 
4751 
9020 

013259 
7451 

021003 
6715 
9789 

033820 
7825 


110 
1 
2 

3 

4 
6 

0 
7 
S 
9 


041393 
6323 
9218 

053078, 
0905 

0G0098 
4458 
8180 

071882 
6547 


041787 
5714 
9000 

053403 
7280 

001075 
4832 
8557 

072250 
5912 


000808 
6181 
9451 

013080 
7808 

022010 
0125 

030195 
4227 
8223 


001301 
5009 
9876 

014100 
8284 

022428 
0533 

030000 
4028 
8020 


001734 
6038 

010300 
4521 
87CT, 

02284 1 
0942 

031004 
5029 
9017 


042182 
0105 
9993 

053840 
7066 

001452 
6200 
8928 

072017 
0270 


042570 
6495 

050380 
4230 
8040 

061829 
5580 
9298 

072985 
6640 


042909 
0885 

0507C6 

,  4013 
8426 

002200 
5953 
9008 

073352 
7004 


002100 
6406 

010724 
4940 
9110 

023252 
7350 

031408 
5430 
9414 


002598 

6894 

011147 

5300 

9532 

023004 

7757 

031812 

6830 

9811 


003029 
7321 

011570 
5779 
9947 

024075 
8164 

032216 
6230 

040207 


003401 
7748 

011993 
6197 

020301 
4480 
8571 

032619 
6629 

040002 


003891 432 
8174428 

012415'424 
00l0i420 

020775416 
4896  412 
8978408 


033021 

7028 

040998 


404 
400 
397 


120 
1 

2 
3 


079181 

082785 

6360 

9905 
4  093422 
5,  6910 
6  100371 
7-  3S04 
8'  7210 
91 10590 


079543 

083144 
6710 

090258 
3772 
7257 

100715 
4146 
7549 

110926 


079904 

083503 
7071 

09001 i 
4122 
7004 

101059 
4487 
7888 

111203 


080206 
3801 
7420 

090903 
4471 
7951 

101403 
4828 
8227 

111599 


0801.20 

080987 

4219 

4576 

7781 

8136 

091315 

091007 

4820 

6109 

8298 

8644 

101747 

102091 

6109 

5510 

8505 

8903 

111934 

112270 

1149441 
8265 

121560! 
4830J 
8076J 

131298 
4496J 
7071 

140822! 
3951 


043302 
7275 
f 51153 

4990 
8805 

002582 
6326 

070038 
3718 
7308 


043755 

7664 
051538 
5378 
9185 
062958 
6699 
070407 
4085 
7731 


044148 
8053 

051924 
6760 
9563 

003333 
7071 

070770 
4451 
8094 


081347 
4934 
8490 

092018 
5518 
8990 

102434 
6851 
9241 

112005 


044540 
8442 

052309 
0142 
9942 

063709 
7443 

071145 
4816 
8457 


081707 
6291 
8845 

092370 
6866 
9335 

102777 
6191 
9579 

112340 


082007 
5047 
9198 

092721 
6215 
9681 

103119 
6531 
9916 

113275 


044932393 
88301390 


052694 

6524 
060320 

4083 

7815 
0716141370 

5182366 

8819363 


380 
383 
379 
376 
373 


0824201360 
6004 


9552 
093071 

6502 
100020 

3462 

6871 
110253 

3  CO  9 


357 
355 
352 
349 
346 
343 
341 
338 
335 


130,113943 

1 

7271 

2 

120574 

3 

3852 

4 

7105 

5 

130334 

6 

3539 

7 

6721 

8 

9879 

9 

143015 

114277 
7003 

120903 
4178 
7429 

130055 
3858 
7037 

140194 
3327 


114011 
7934 

121231 
4504 
7753 

130977 
4177 
7354 

140508 
3039 


115278 
8595 

121888 
5156 
8399 

131019 
4814 
7987 

141136 
4203 


115011 
8926 

122216 
5481 
8722 

131939 
5133 
8303 

141450 
4574 


115943 
9256 

122544 
5800 
9045 

132200 
5451 
8018 

141703 
4885 


116276 
9586 

122871 
6131 
9308 

132580 
5769 
8934 

142070 
5190 


116608 
9915 

123198 
6456 
9090 

132900 
6080 
9249 

142389 
5507 


116940333 
1-20245  330 
3525  328 
6781325 
130012  323 
3219  321 
6403i318 


9504 

142702 

5818 


310 
314 
311 


140 
11 
2 
3 
4 
5 
6 
7 
8 
9 

150 
1 
2 
3 
4 
5 
6 
7 
8 
9 

NTT 


140128 
9219 

152288 
6330 
8302 

101308 
4353 
7317 

170202 
3180 

176091 
8977 

181844 
4091 

7521 
190332 

3125 
5900 
8057 
201397 
0 


146438 
9527 

152594 
5640 
8664 

101007 
4050 
7013 

170555 
3478 

176381 

9204 

182129 

4975 

7803 

190012 

3403 

0170 

8932 

201070 

i 


140748 
9835 

152900 
5.U3 
8005 

101907 
4947 
7908 

170848 
3709 


,147058 

150142 
3205 
0240 
9206 

1 02200 
5244 
8203 

171141 
4000, 


170070 
9552 

182415 
62591 
8084 1 

190892 

31.81 

0153 

9206 

201913 


176959 
9839 

182700 
6542 
8306 

191171 
3959 
0729 
9481 

202216 

3   I 


147307 

150449 
3510 
0549 
9507 

102504 
5541 
8497 

171434 
4351 

177248 

180120 
2985 
6825 
8047 

191451 
4  237 
7005 
9755 

2(1 2  1 88 


147070 

'S0750 
3815 
3852 
9808 

102803 
5838 
8792 

171720 
4641 

177530 

180413 
3270 
0108 
8928 

191730 
4514 
7281 

200029 
2761 


II 


147985 

151003 
4120 
7154 

160108 
3101 
6134 
9086 

172019 
4H32 

177825 

180099 
3555 
639J 
9209 

192010 
4792 
7556 

200303 
3033 
o 


148291 

151370 
4424 
7457 

100409 
3400 
6430 
9380 

172311 
5222 

178113 

1801180 
3839 
6674 
9490 

19228H 
6069 
7832 

1200577 
3305 


148003 

151670 
4728 
7759 

100709 
3758 
0720 
9074 

172003 
6512 


148911 

151982 
5032 
f-001 

161008 
4055 
7022 
9908 

172895 
6802 


309 

307 

305 

303, 

30ll 

299- 

297i 

295 

293 

291 


178401 

181272 
4123 
6956 
9771 

192567 
6346 
8107 

200850 
3577 


178689  289 

181558  287 
4407  285 
7239.283 

190051 '281 
2840  279 
8623  278 
8382276 

201124  271 
3818  272 
T~Ti  |  n 


OF    NUMBERS. 


N.|   0  |   ]     2     3 

4 

5 

6     7     8     9   ,D. 

160  204120  204391 

204663 

204934 

205204 

205475 

205746 

206016 206286 

206556)271 

1 

G82G 

709G 

73G5 

7G34 

7904 

8173 

8441 

8710 

8979 

9247  269 

2 

9515 

9783 

210051 

210319 

210586 

210853 

211121 

211388 

211654 

211921'2C7 

3 

212188  212454 

2720 

2986 

3252 

3518 

3783 

4049 

4314 

4579  266 

4 

4844 

6109 

6373 

6638 

5902 

6166 

6430 

6694 

6957 

7221 

264 

5 

7484 

7747 

8010 

8273 

8536 

8798 

9060 

9323 

9585 

9846 

262 

6 

220108 

220370 

220631 

220892 

221153 

221414 

221675  221936 

222196 

222456 

261 

7 

2716 

297G 

3236 

3496 

3755 

4015 

4274 

4533 

4792 

6051 

259 

8 

5309 

5568 

5826 

6084 

6342 

6600 

6858 

7115 

7372 

7630 

258 

9 

7887 

8144 

8400 

8057 

231215 

8913 
231470 

9170 

231724 

9426 
23T9~79 

9G82 
232234 

9938  230193 

256 

170 

230449  230704 

230960 

232488  232742 

255 

1 

2996   3250 

3504 

3757 

4011 

4264 

4517 

4770 

6023 

6276 

253 

2 

5528   5781 

6033 

6285 

6537 

6789 

7041 

7292 

7544 

7795 

252 

3 

804G1  8297 

8548 

8799 

9049 

9299 

9550 

9800 

240050 

240300 

250 

4 

240549  240799 

241048 

241297 

241546 

241795  242044 

242293 

2541 

2790 

249 

6 

3038 

3286 

3534 

3782 

4030 

4277 

4525 

4772 

5019 

6266 

248 

6 

5513 

5759 

6006 

6252 

6499 

6745 

6991 

7237 

7482 

7728 

246 

7 

7973 

8219 

84G4 

8709 

8954 

9198 

9443 

9687 

9932 

250176 

•245 

8 

250420 

250GG4 

250908 

251151  251395 

251638 

251881 

252125 

252368 

2610 

24o 

9 

2853 

3096 
255514 

3338 
255755 

3580|  3822 

4064 

4306 
256718 

4548 

4790 

5031 

242 

180 

255273 

255996 

256237 

256477 

256958 

257198 

257439 

241 

1 

7679 

7918 

8158 

8398 

8637 

8877 

9116 

9355 

9594 

9833 

239 

2 

260071 

260310 

260548 

260787 

261025 

261263 

261501 

261739 

261976 

262214 

238 

3 

2451 

2G88 

2925 

3162 

3399 

3G3G 

3873 

4109 

4346 

4582 

237 

4 

4818 

5054 

6290 

5525 

6761 

6996 

6232 

6467 

6702 

G037 

235 

6 

7172 

7406 

7641 

7875 

8110 

8344 

8578 

8812 

904G 

9279 

234 

6 

9513 

9746 

9980 

270213 

270446 

270679 

270912 

271144 

271377 

271609 

233 

1 

271842 

272074 

272306 

2538 

2770 

3001 

3233 

3464 

3696 

3927 

232 

8 

4158 

4389 

4620 

4850 

5081 

5311 

6542 

6772 

6002 

6232 

230 

9 

190 

6462 

278754 

6692 

6921 

7151 

7380 

7609 

7838 

8067 

8296 

8525 

229 

278982:279211 

2794391279667, 

281715281942 

279895  28 

280351  21 

228 

1 

281033 

281261:281488 

282169 

2396 

2622 

2849 

3075 

227 

2 

3301 

3527  3753 

3979   4205 

4431 

4656 

4882 

5107 

5332 

226 

3 

6557 

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218 

200  301030 

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197 

1 

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2 

6353 

6549 

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3 

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4 

350248 

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351410 

351(03 

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193 

6 

2183 

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3147 

3339 

3532 

3724 

3916 

103 

6 

4108 

4301 

4493 

4685 

4876 

6068 

5260 

6452 

6643 

5834 

102 

7 

6026 

6217 

6408 

6599 

6790 

6981 

7172 

7363 

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7744 

101 

8 

7935 

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8316 

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8886 

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9266 

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9646 

190 

9 

9835 

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360215  360404 

360593 

360783 

360972 

361161 

361350 

361539 

189 

230 

361728 

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1 

3612 

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2 

6488 

6675 

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6049 

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6610 

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187 

3 

7356 

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8845 

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180 

4 

9216 

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370513 

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185 

5 

371068 

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2175 

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184 

6 

2912 

3096 

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3647 

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7 

4748 

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5664 

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181 

240 

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181 

1 

2017 

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180 

2 

3815 

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4712   4891 

6070 

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179 

3 

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7034 

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178 

4 

7390 

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8634 

8811 

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178 

5 

9166 

9343 

9520 

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9875 

300051 

390228 

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390582  390759 

177 

6 

390935 

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391464 

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176 

7 

2697 

2873 

3048 

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3400 

3575 

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4101   4277 

176 

8 

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5501 

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5850   6025 

175 

9 

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7592|  7766 

174 

250 

397940 

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398461  398634 

308808  398981,399154 

399328  399501 

173 

1 

9674 

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400192  400365 

400538  4007111400883 

401056 

401228  173 

2 

401401 

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1745 

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2089 

2261 

24331  2605 

2777 

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172 

3 

3121 

3292 

3464   3635 

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4149   4320 

4492 

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171 

4 

4834 

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6 

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411620 

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163 

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9 

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431364 

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432640  432809 

161 

1 

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2 

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3   6163 

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7 

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5 

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146 

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LOGARITHMS 

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3989   40381  4088 

4137 

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42361  4285 

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43841  4433  49 

N. 

0 

1   |   2 

3 

4   II 

6 

7 

8     9   ID. 

OF    NUMBERS 


15 


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1229 

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9  .d. 

16 


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nTT 

1  o 

1 

2 

4 

J  & 

6 

7 

8 

9   ID. 

14  DAY  USE 

RETURN  TO  DESK  FROM  WHT     t  BORROWED 


LOAN  D 


This  book  is  due  on  the  la- 
or  on  the  date  to  which  r 
B„no      .  Tel.  No. 

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